Codeforces Global Round 6
阿新 • • 發佈：2019-12-21
久違的寫篇部落格吧
A. Competitive Programmer
題目連結：https://codeforces.com/contest/1266/problem/A
題意：
給你一個只包含數字的字串，你可以將字串隨機排列，問最後組成的數字能否被60整除
分析：
我們先考慮能被6整除的數的特點：
①各個位數和 % 3 == 0
②末尾數必須為 0 2 4 6 8中的一個
再考慮能被10整除的數的特點：
末尾數必須為0
因為60中6在十位部分 ，10在個位部分，所以最後組成的數除了各個位數和 % 3 == 0 外十位部分必須為0 、2 、4、6、8 中的一個， 個位數必須為0。
#include<bits/stdc++.h>
#define ios std::ios::sync_with_stdio(false)
#define sd(n) scanf("%d",&n)
#define sdd(n,m) scanf("%d%d",&n,&m)
#define sddd(n,m,k) scanf("%d%d%d",&n,&m,&k)
#define pd(n) printf("%d\n", (n))
#define pdd(n,m) printf("%d %d\n", n, m)
#define pld(n) printf("%lld\n", n)
#define pldd(n,m) printf("%lld %lld\n", n, m)
#define sld(n) scanf("%lld",&n)
#define sldd(n,m) scanf("%lld%lld",&n,&m)
#define slddd(n,m,k) scanf("%lld%lld%lld",&n,&m,&k)
#define sf(n) scanf("%lf",&n)
#define sff(n,m) scanf("%lf%lf",&n,&m)
#define sfff(n,m,k) scanf("%lf%lf%lf",&n,&m,&k)
#define rep(i,a,n) for (int i=a;i<=n;i++)
#define per(i,n,a) for (int i=n;i>=a;i--)
#define mm(a,n) memset(a, n, sizeof(a))
#define pb push_back
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define il inline
#define ll long long
#define lson rt << 1
#define rson rt << 1 | 1
#define MOD 1000000007
#define pi 3.14159265358979323
#define debug(x) cout <<#x<<": "<<x<<endl
#define debug2(x, y) cout <<#x<<": "<<x<<" | "<<#y<<": "<<y<<endl;
#define debug3(x, y, z) cout <<#x<<": "<<x<<" | "<<#y<<": "<<y<<" | "<<#z<<": "<<z<<endl;
#define debug4(a, b, c, d) cout <<#a<<": "<<a<<" | "<<#b<<": "<<b<<" | "<<#c<<": "<<c<<" | "<<#d<<": "<<d<<endl;
using namespace std;
const ll INF (0x3f3f3f3f3f3f3f3fll);
const int inf (0x3f3f3f3f);
template<typename T>void read(T &res){bool flag=false;char ch;while(!isdigit(ch=getchar()))(ch=='-')&&(flag=true);
for(res=ch-48;isdigit(ch=getchar());res=(res<<1)+(res<<3)+ch - 48);flag&&(res=-res);}
template<typename T>void Out(T x){if(x<0)putchar('-'),x=-x;if(x>9)Out(x/10);putchar(x%10+'0');}
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a*b/gcd(a,b);}
ll pow_mod(ll x,ll n,ll mod){ll res=1;while(n){if(n&1)res=res*x%mod;x=x*x%mod;n>>=1;}return res;}
ll fact_pow(ll n,ll p){ll res=0;while(n){n/=p;res+=n;}return res;}
ll mult(ll a,ll b,ll p){a%=p;b%=p;ll r=0,v=a;while(b){if(b&1){r+=v;if(r>p)r-=p;}v<<=1;if(v>p)v-=p;b>>=1;}return r;}
ll quick_pow(ll a,ll b,ll p){ll r=1,v=a%p;while(b){if(b&1)r=mult(r,v,p);v=mult(v,v,p);b>>=1;}return r;}
bool CH(ll a,ll n,ll x,ll t)
{ll r=quick_pow(a,x,n);ll z=r;for(ll i=1;i<=t;i++){r=mult(r,r,n);if(r==1&&z!=1&&z!=n-1)return true;z=r;}return r!=1;}
bool Miller_Rabin(ll n)
{if(n<2)return false;if(n==2)return true;if(!(n&1))return false;ll x=n-1,t=0;while(!(x&1)){x>>=1;t++;}srand(time(NULL));
ll o=8;for(ll i=0;i<o;i++){ll a=rand()%(n-1)+1;if(CH(a,n,x,t))return false;}return true;}
/*
int prime[30000010],minprime[30000010];
void euler(int n)
{int c=0,i,j;for(i=2;i<=n;i++){if(!minprime[i])prime[++c]=i,minprime[i]=i;for(j=1;j<=c&&i*prime[j]<=n;j++)
{minprime[i*prime[j]]=prime[j];if(i%prime[j]==0)break;}}}
struct Tree{ll l,r,sum,lazy,maxn,minn;}tree[1000000];
il void push_up(ll rt)
{tree[rt].sum=tree[rt<<1].sum+tree[rt<<1|1].sum;
tree[rt].maxn=max(tree[rt<<1].maxn,tree[rt<<1|1].maxn);
tree[rt].minn=min(tree[rt<<1].minn,tree[rt<<1|1].minn);}
il void push_down(ll rt , ll now)
{if(tree[rt].lazy){tree[rt<<1].lazy+=tree[rt].lazy;tree[rt<<1|1].lazy+=tree[rt].lazy;
tree[rt<<1].sum+=(now-(now>>1))*tree[rt].lazy;tree[rt<<1|1].sum+=(now>>1)*tree[rt].lazy;
tree[rt<<1].minn+=tree[rt].lazy;tree[rt<<1|1].minn+=tree[rt].lazy;
tree[rt<<1].maxn+=tree[rt].lazy;tree[rt<<1|1].maxn+=tree[rt].lazy;tree[rt].lazy=0;}}
il void build(ll l , ll r , ll rt , ll *aa)
{tree[rt].lazy=0;tree[rt].l=l;tree[rt].r=r;if(l==r)
{tree[rt].sum=aa[l];tree[rt].minn=tree[rt].sum;tree[rt].maxn=tree[rt].sum;return;}
ll mid=(l+r)>>1;build(l,mid,rt<<1,aa);build(mid+1,r,rt<<1|1,aa);push_up(rt);}
il void update_range(ll L , ll R , ll key , ll rt)
{if(L<=tree[rt].l&&R>=tree[rt].r){tree[rt].sum+=(tree[rt].r-tree[rt].l+1)*key;
tree[rt].minn+=key;tree[rt].maxn+=key;tree[rt].lazy+=key;return;}push_down(rt,tree[rt].r-tree[rt].l+1);
ll mid=(tree[rt].r+tree[rt].l)>>1;if(L<=mid)update_range(L,R,key,lson);
if(R>mid)update_range(L,R,key,rson);push_up(rt);}
il ll query_range(ll L, ll R, ll rt)
{if(L<=tree[rt].l&&R>=tree[rt].r){return tree[rt].sum;}push_down(rt,tree[rt].r-tree[rt].l+1);
ll mid=(tree[rt].r+tree[rt].l)>>1;ll ans=0;if(L<=mid)ans+=query_range(L,R,lson);
if(R>mid)ans+=query_range(L,R,rson);return ans;}
il ll query_min(ll L,ll R,ll rt)
{if(L<=tree[rt].l&&R>=tree[rt].r){return tree[rt].minn;}push_down(rt,tree[rt].r-tree[rt].l+1);
ll mid=(tree[rt].r+tree[rt].l)>>1;ll ans=(0x3f3f3f3f3f3f3f3fll);if(L<=mid)ans=min(ans,query_min(L,R,lson));
if(R>mid)ans=min(ans,query_min(L,R,rson));return ans;}
il ll query_max(ll L, ll R, ll rt)
{if(L<=tree[rt].l&&R>=tree[rt].r){return tree[rt].maxn;}push_down(rt,tree[rt].r-tree[rt].l+1);
ll mid=(tree[rt].r+tree[rt].l)>>1;ll ans=-(0x3f3f3f3f3f3f3f3fll);if(L<=mid)ans=max(ans,query_max(L,R,lson));
if(R>mid)ans=max(ans,query_max(L,R,rson));return ans;}
*/
const int N = 1e3 + 10;
ll a[N][N];
map<int , int>ha;
int main()
{
ios;
int n ;
cin >> n;
while(n --)
{
ha.clear();
string s;
cin >> s;
int len = s.size() - 1;
int tot = 0;
rep(j , 0 , len)
{
ha[s[j] - '0'] ++;
tot += s[j] - '0';
}
if(ha[0] && (ha[2] || ha[4] || ha[6] || ha[8] || (ha[0] - 1)) && tot % 3 == 0)
cout << "red" << '\n';
else cout << "cyan" << '\n';
}
return 0;
}
B. Dice Tower
題目連結：https://codeforces.com/contest/1266/problem/B
題意：
你有無限個骰子，你可以從中選出任意個來拼搭，問“沒有被擋住的點數的和”能否組成某個數X
分析：
首先一個骰子所有點數和為21 ， 並且1的對立面為6，2的對立面為5，3對立面為4。
當骰子只有一個的時候，它被擋住的只會它的底部，那麼它可以組成的數就為21 - （1~6） = 15 ~ 20
當骰子有兩個的時候，因為它們要拼搭在一起，所以第一個骰子會被第二個骰子擋住一個面，第二個骰子會被第一個骰子擋住一個面，並且這兩個面是對立的
所以當第二個骰子加入的時候，總的點數為原先點數 + 21（第二個骰子的總點數） - 7（第二個骰子加入時減去的部分） = （15 ~ 20） + 14
加入第三個第四個第五個。。。第n個同理 ， 最後的總點數就為 （15 ~ 20） + 14 * (n - 1) ， 所以如果可以拼湊出X則 X % 14 的答案為 （15 ~ 20 ）% 14 = 1 ~ 6
所以最後只要判斷X是否大於等於15並且 X % 14 在不在 1 ~ 6 這個範圍就可以了
#include<bits/stdc++.h>
#define ios std::ios::sync_with_stdio(false)
#define sd(n) scanf("%d",&n)
#define sdd(n,m) scanf("%d%d",&n,&m)
#define sddd(n,m,k) scanf("%d%d%d",&n,&m,&k)
#define pd(n) printf("%d\n", (n))
#define pdd(n,m) printf("%d %d\n", n, m)
#define pld(n) printf("%lld\n", n)
#define pldd(n,m) printf("%lld %lld\n", n, m)
#define sld(n) scanf("%lld",&n)
#define sldd(n,m) scanf("%lld%lld",&n,&m)
#define slddd(n,m,k) scanf("%lld%lld%lld",&n,&m,&k)
#define sf(n) scanf("%lf",&n)
#define sff(n,m) scanf("%lf%lf",&n,&m)
#define sfff(n,m,k) scanf("%lf%lf%lf",&n,&m,&k)
#define rep(i,a,n) for (int i=a;i<=n;i++)
#define per(i,n,a) for (int i=n;i>=a;i--)
#define mm(a,n) memset(a, n, sizeof(a))
#define pb push_back
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define il inline
#define ll long long
#define lson rt << 1
#define rson rt << 1 | 1
#define MOD 1000000007
#define pi 3.14159265358979323
#define debug(x) cout <<#x<<": "<<x<<endl
#define debug2(x, y) cout <<#x<<": "<<x<<" | "<<#y<<": "<<y<<endl;
#define debug3(x, y, z) cout <<#x<<": "<<x<<" | "<<#y<<": "<<y<<" | "<<#z<<": "<<z<<endl;
#define debug4(a, b, c, d) cout <<#a<<": "<<a<<" | "<<#b<<": "<<b<<" | "<<#c<<": "<<c<<" | "<<#d<<": "<<d<<endl;
using namespace std;
const ll INF (0x3f3f3f3f3f3f3f3fll);
const int inf (0x3f3f3f3f);
template<typename T>void read(T &res){bool flag=false;char ch;while(!isdigit(ch=getchar()))(ch=='-')&&(flag=true);
for(res=ch-48;isdigit(ch=getchar());res=(res<<1)+(res<<3)+ch - 48);flag&&(res=-res);}
template<typename T>void Out(T x){if(x<0)putchar('-'),x=-x;if(x>9)Out(x/10);putchar(x%10+'0');}
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a*b/gcd(a,b);}
ll pow_mod(ll x,ll n,ll mod){ll res=1;while(n){if(n&1)res=res*x%mod;x=x*x%mod;n>>=1;}return res;}
ll fact_pow(ll n,ll p){ll res=0;while(n){n/=p;res+=n;}return res;}
ll mult(ll a,ll b,ll p){a%=p;b%=p;ll r=0,v=a;while(b){if(b&1){r+=v;if(r>p)r-=p;}v<<=1;if(v>p)v-=p;b>>=1;}return r;}
ll quick_pow(ll a,ll b,ll p){ll r=1,v=a%p;while(b){if(b&1)r=mult(r,v,p);v=mult(v,v,p);b>>=1;}return r;}
bool CH(ll a,ll n,ll x,ll t)
{ll r=quick_pow(a,x,n);ll z=r;for(ll i=1;i<=t;i++){r=mult(r,r,n);if(r==1&&z!=1&&z!=n-1)return true;z=r;}return r!=1;}
bool Miller_Rabin(ll n)
{if(n<2)return false;if(n==2)return true;if(!(n&1))return false;ll x=n-1,t=0;while(!(x&1)){x>>=1;t++;}srand(time(NULL));
ll o=8;for(ll i=0;i<o;i++){ll a=rand()%(n-1)+1;if(CH(a,n,x,t))return false;}return true;}
/*
int prime[30000010],minprime[30000010];
void euler(int n)
{int c=0,i,j;for(i=2;i<=n;i++){if(!minprime[i])prime[++c]=i,minprime[i]=i;for(j=1;j<=c&&i*prime[j]<=n;j++)
{minprime[i*prime[j]]=prime[j];if(i%prime[j]==0)break;}}}
struct Tree{ll l,r,sum,lazy,maxn,minn;}tree[1000000];
il void push_up(ll rt)
{tree[rt].sum=tree[rt<<1].sum+tree[rt<<1|1].sum;
tree[rt].maxn=max(tree[rt<<1].maxn,tree[rt<<1|1].maxn);
tree[rt].minn=min(tree[rt<<1].minn,tree[rt<<1|1].minn);}
il void push_down(ll rt , ll now)
{if(tree[rt].lazy){tree[rt<<1].lazy+=tree[rt].lazy;tree[rt<<1|1].lazy+=tree[rt].lazy;
tree[rt<<1].sum+=(now-(now>>1))*tree[rt].lazy;tree[rt<<1|1].sum+=(now>>1)*tree[rt].lazy;
tree[rt<<1].minn+=tree[rt].lazy;tree[rt<<1|1].minn+=tree[rt].lazy;
tree[rt<<1].maxn+=tree[rt].lazy;tree[rt<<1|1].maxn+=tree[rt].lazy;tree[rt].lazy=0;}}
il void build(ll l , ll r , ll rt , ll *aa)
{tree[rt].lazy=0;tree[rt].l=l;tree[rt].r=r;if(l==r)
{tree[rt].sum=aa[l];tree[rt].minn=tree[rt].sum;tree[rt].maxn=tree[rt].sum;return;}
ll mid=(l+r)>>1;build(l,mid,rt<<1,aa);build(mid+1,r,rt<<1|1,aa);push_up(rt);}
il void update_range(ll L , ll R , ll key , ll rt)
{if(L<=tree[rt].l&&R>=tree[rt].r){tree[rt].sum+=(tree[rt].r-tree[rt].l+1)*key;
tree[rt].minn+=key;tree[rt].maxn+=key;tree[rt].lazy+=key;return;}push_down(rt,tree[rt].r-tree[rt].l+1);
ll mid=(tree[rt].r+tree[rt].l)>>1;if(L<=mid)update_range(L,R,key,lson);
if(R>mid)update_range(L,R,key,rson);push_up(rt);}
il ll query_range(ll L, ll R, ll rt)
{if(L<=tree[rt].l&&R>=tree[rt].r){return tree[rt].sum;}push_down(rt,tree[rt].r-tree[rt].l+1);
ll mid=(tree[rt].r+tree[rt].l)>>1;ll ans=0;if(L<=mid)ans+=query_range(L,R,lson);
if(R>mid)ans+=query_range(L,R,rson);return ans;}
il ll query_min(ll L,ll R,ll rt)
{if(L<=tree[rt].l&&R>=tree[rt].r){return tree[rt].minn;}push_down(rt,tree[rt].r-tree[rt].l+1);
ll mid=(tree[rt].r+tree[rt].l)>>1;ll ans=(0x3f3f3f3f3f3f3f3fll);if(L<=mid)ans=min(ans,query_min(L,R,lson));
if(R>mid)ans=min(ans,query_min(L,R,rson));return ans;}
il ll query_max(ll L, ll R, ll rt)
{if(L<=tree[rt].l&&R>=tree[rt].r){return tree[rt].maxn;}push_down(rt,tree[rt].r-tree[rt].l+1);
ll mid=(tree[rt].r+tree[rt].l)>>1;ll ans=-(0x3f3f3f3f3f3f3f3fll);if(L<=mid)ans=max(ans,query_max(L,R,lson));
if(R>mid)ans=max(ans,query_max(L,R,rson));return ans;}
*/
const int N = 1e3 + 10;
ll a[N][N];
map<int , int>ha;
int main()
{
ios;
int t;
cin >> t;
while(t --)
{
ll x;
cin >> x;
if(x < 15)
{
cout << "NO" << '\n';
continue;
}
x %= 14;
if(x > 6 || x < 1)
cout << "NO" << '\n';
else
cout << "YES" << '\n';
}
return 0;
}
C. Diverse Matrix
題目連結：https://codeforces.com/contest/1266/problem/C
題意：
要求構造一個 r 行 c 列的矩陣， 並且矩陣的每一行的gcd和每一列的gcd的都只出現過一次，且每行的gcd都要求最小
分析：
首先當 r = 1 && c = 1 的時候， 它的行gcd和列gcd為同一元素， 所以是無法構造出來的
當 r = 1 或者 c = 1 時 ， 我們可以 1 2 3 4 5 6 ... 輸出
對於其它的可能，我們考慮這麼構造：
假設 r = 2 ， c = 2 ， A 、B 、C 、D 為矩陣元素
1 | A C
2 | B D 我們可以讓第一行的gcd結果為1 ， 第二行gcd結果為2，第一列gcd結果為3，第二列gcd結果為4
3 4
則 A = 1 * 3 , B = 2 * 3 , C = 1 * 4 , D = 2 * 4 這樣構造即可
#include<bits/stdc++.h>
#define ios std::ios::sync_with_stdio(false)
#define sd(n) scanf("%d",&n)
#define sdd(n,m) scanf("%d%d",&n,&m)
#define sddd(n,m,k) scanf("%d%d%d",&n,&m,&k)
#define pd(n) printf("%d\n", (n))
#define pdd(n,m) printf("%d %d\n", n, m)
#define pld(n) printf("%lld\n", n)
#define pldd(n,m) printf("%lld %lld\n", n, m)
#define sld(n) scanf("%lld",&n)
#define sldd(n,m) scanf("%lld%lld",&n,&m)
#define slddd(n,m,k) scanf("%lld%lld%lld",&n,&m,&k)
#define sf(n) scanf("%lf",&n)
#define sff(n,m) scanf("%lf%lf",&n,&m)
#define sfff(n,m,k) scanf("%lf%lf%lf",&n,&m,&k)
#define rep(i,a,n) for (int i=a;i<=n;i++)
#define per(i,n,a) for (int i=n;i>=a;i--)
#define mm(a,n) memset(a, n, sizeof(a))
#define pb push_back
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define il inline
#define ll long long
#define lson rt << 1
#define rson rt << 1 | 1
#define MOD 1000000007
#define pi 3.14159265358979323
#define debug(x) cout <<#x<<": "<<x<<endl
#define debug2(x, y) cout <<#x<<": "<<x<<" | "<<#y<<": "<<y<<endl;
#define debug3(x, y, z) cout <<#x<<": "<<x<<" | "<<#y<<": "<<y<<" | "<<#z<<": "<<z<<endl;
#define debug4(a, b, c, d) cout <<#a<<": "<<a<<" | "<<#b<<": "<<b<<" | "<<#c<<": "<<c<<" | "<<#d<<": "<<d<<endl;
using namespace std;
const ll INF (0x3f3f3f3f3f3f3f3fll);
const int inf (0x3f3f3f3f);
template<typename T>void read(T &res){bool flag=false;char ch;while(!isdigit(ch=getchar()))(ch=='-')&&(flag=true);
for(res=ch-48;isdigit(ch=getchar());res=(res<<1)+(res<<3)+ch - 48);flag&&(res=-res);}
template<typename T>void Out(T x){if(x<0)putchar('-'),x=-x;if(x>9)Out(x/10);putchar(x%10+'0');}
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a*b/gcd(a,b);}
ll pow_mod(ll x,ll n,ll mod){ll res=1;while(n){if(n&1)res=res*x%mod;x=x*x%mod;n>>=1;}return res;}
ll fact_pow(ll n,ll p){ll res=0;while(n){n/=p;res+=n;}return res;}
ll mult(ll a,ll b,ll p){a%=p;b%=p;ll r=0,v=a;while(b){if(b&1){r+=v;if(r>p)r-=p;}v<<=1;if(v>p)v-=p;b>>=1;}return r;}
ll quick_pow(ll a,ll b,ll p){ll r=1,v=a%p;while(b){if(b&1)r=mult(r,v,p);v=mult(v,v,p);b>>=1;}return r;}
bool CH(ll a,ll n,ll x,ll t)
{ll r=quick_pow(a,x,n);ll z=r;for(ll i=1;i<=t;i++){r=mult(r,r,n);if(r==1&&z!=1&&z!=n-1)return true;z=r;}return r!=1;}
bool Miller_Rabin(ll n)
{if(n<2)return false;if(n==2)return true;if(!(n&1))return false;ll x=n-1,t=0;while(!(x&1)){x>>=1;t++;}srand(time(NULL));
ll o=8;for(ll i=0;i<o;i++){ll a=rand()%(n-1)+1;if(CH(a,n,x,t))return false;}return true;}
/*
int prime[30000010],minprime[30000010];
void euler(int n)
{int c=0,i,j;for(i=2;i<=n;i++){if(!minprime[i])prime[++c]=i,minprime[i]=i;for(j=1;j<=c&&i*prime[j]<=n;j++)
{minprime[i*prime[j]]=prime[j];if(i%prime[j]==0)break;}}}
struct Tree{ll l,r,sum,lazy,maxn,minn;}tree[1000000];
il void push_up(ll rt)
{tree[rt].sum=tree[rt<<1].sum+tree[rt<<1|1].sum;
tree[rt].maxn=max(tree[rt<<1].maxn,tree[rt<<1|1].maxn);
tree[rt].minn=min(tree[rt<<1].minn,tree[rt<<1|1].minn);}
il void push_down(ll rt , ll now)
{if(tree[rt].lazy){tree[rt<<1].lazy+=tree[rt].lazy;tree[rt<<1|1].lazy+=tree[rt].lazy;
tree[rt<<1].sum+=(now-(now>>1))*tree[rt].lazy;tree[rt<<1|1].sum+=(now>>1)*tree[rt].lazy;
tree[rt<<1].minn+=tree[rt].lazy;tree[rt<<1|1].minn+=tree[rt].lazy;
tree[rt<<1].maxn+=tree[rt].lazy;tree[rt<<1|1].maxn+=tree[rt].lazy;tree[rt].lazy=0;}}
il void build(ll l , ll r , ll rt , ll *aa)
{tree[rt].lazy=0;tree[rt].l=l;tree[rt].r=r;if(l==r)
{tree[rt].sum=aa[l];tree[rt].minn=tree[rt].sum;tree[rt].maxn=tree[rt].sum;return;}
ll mid=(l+r)>>1;build(l,mid,rt<<1,aa);build(mid+1,r,rt<<1|1,aa);push_up(rt);}
il void update_range(ll L , ll R , ll key , ll rt)
{if(L<=tree[rt].l&&R>=tree[rt].r){tree[rt].sum+=(tree[rt].r-tree[rt].l+1)*key;
tree[rt].minn+=key;tree[rt].maxn+=key;tree[rt].lazy+=key;return;}push_down(rt,tree[rt].r-tree[rt].l+1);
ll mid=(tree[rt].r+tree[rt].l)>>1;if(L<=mid)update_range(L,R,key,lson);
if(R>mid)update_range(L,R,key,rson);push_up(rt);}
il ll query_range(ll L, ll R, ll rt)
{if(L<=tree[rt].l&&R>=tree[rt].r){return tree[rt].sum;}push_down(rt,tree[rt].r-tree[rt].l+1);
ll mid=(tree[rt].r+tree[rt].l)>>1;ll ans=0;if(L<=mid)ans+=query_range(L,R,lson);
if(R>mid)ans+=query_range(L,R,rson);return ans;}
il ll query_min(ll L,ll R,ll rt)
{if(L<=tree[rt].l&&R>=tree[rt].r){return tree[rt].minn;}push_down(rt,tree[rt].r-tree[rt].l+1);
ll mid=(tree[rt].r+tree[rt].l)>>1;ll ans=(0x3f3f3f3f3f3f3f3fll);if(L<=mid)ans=min(ans,query_min(L,R,lson));
if(R>mid)ans=min(ans,query_min(L,R,rson));return ans;}
il ll query_max(ll L, ll R, ll rt)
{if(L<=tree[rt].l&&R>=tree[rt].r){return tree[rt].maxn;}push_down(rt,tree[rt].r-tree[rt].l+1);
ll mid=(tree[rt].r+tree[rt].l)>>1;ll ans=-(0x3f3f3f3f3f3f3f3fll);if(L<=mid)ans=max(ans,query_max(L,R,lson));
if(R>mid)ans=max(ans,query_max(L,R,rson));return ans;}
*/
const int N = 1e3 + 10;
ll a[N][N];
int main()
{
ios;
ll r , c;
cin >> r >> c;
if(r == 1 && c == 1)
return cout << 0 << '\n' , 0;
if(r == 1)
{
int tot = 2;
rep(i , 1 , c)
cout << tot ++ << ' ';
return 0;
}
if(c == 1)
{
int tot = 2;
rep(i , 1 , r)
cout << tot ++ << "\n";
return 0;
}
rep(i , 1 , r)
{
rep(j , 1 , c)
{
a[i][j] = i * (j + r);
}
}
rep(i , 1 , r)
{
rep(j , 1 , c)
cout << a[i][j] << " ";
cout << '\n';
}
return 0;
}
View Code
D. Decreasing Debts
題目連結：https://codeforces.ml/contest/1266/problem/D
題意：
給你幾個債務關係，如A欠B五元， B欠C五元，此時總的債務為 5 + 5 = 10 。
我們可以把關係轉換為A欠C五元，那這樣總的債務為 5
問不限轉換次數 ， 怎麼把總債務化為最小。
分析：
當把總債務化為最小時 ， 每個人的債務關係也將是最簡的（以上述例子闡述，對B , 它欠C五元，這時候它的支出為5，但是A欠它五元，所以它的收入為5，所以它最後對總債務的關係為abs（5 - 5）= 0）
對於A、C也是一樣的，所以最後A的最簡形式為A支出5元,B的最簡形式為0，C的最簡形式為收入5元。按照這樣，我們只要將A指向C即可
（因為總的收入和總的支出一定是相同的且題目也說債務關係數量最小，所以即使某個人 D 收入的金額大於另一個人 E 支出的金額，也可以先將E連向D，然後D的收入金額減去E的支出金額）
我們用 in 容器來存最簡形式為收入的人的編號， ou容器來存最簡形式為支出的人的編號，再用指標one指向in裡的元素，two指向ou裡的元素，他們的債務可以表示為min(a[one] , a[two]) , 將關係存於ans容器裡。
當one == in.size() || two == ou.size() 即已經沒有支出的人或已經沒有收入的人了，則債務關係可以結束儲存（題目要求最後輸出債務關係）
#include<bits/stdc++.h>
#define ios std::ios::sync_with_stdio(false)
#define sd(n) scanf("%d",&n)
#define sdd(n,m) scanf("%d%d",&n,&m)
#define sddd(n,m,k) scanf("%d%d%d",&n,&m,&k)
#define pd(n) printf("%d\n", (n))
#define pdd(n,m) printf("%d %d\n", n, m)
#define pld(n) printf("%lld\n", n)
#define pldd(n,m) printf("%lld %lld\n", n, m)
#define sld(n) scanf("%lld",&n)
#define sldd(n,m) scanf("%lld%lld",&n,&m)
#define slddd(n,m,k) scanf("%lld%lld%lld",&n,&m,&k)
#define sf(n) scanf("%lf",&n)
#define sff(n,m) scanf("%lf%lf",&n,&m)
#define sfff(n,m,k) scanf("%lf%lf%lf",&n,&m,&k)
#define rep(i,a,n) for (int i=a;i<=n;i++)
#define per(i,n,a) for (int i=n;i>=a;i--)
#define mm(a,n) memset(a, n, sizeof(a))
#define pb push_back
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define il inline
#define ll long long
#define lson rt << 1
#define rson rt << 1 | 1
#define MOD 1000000007
#define pi 3.14159265358979323
#define debug(x) cout <<#x<<": "<<x<<endl
#define debug2(x, y) cout <<#x<<": "<<x<<" | "<<#y<<": "<<y<<endl;
#define debug3(x, y, z) cout <<#x<<": "<<x<<" | "<<#y<<": "<<y<<" | "<<#z<<": "<<z<<endl;
#define debug4(a, b, c, d) cout <<#a<<": "<<a<<" | "<<#b<<": "<<b<<" | "<<#c<<": "<<c<<" | "<<#d<<": "<<d<<endl;
using namespace std;
const ll INF (0x3f3f3f3f3f3f3f3fll);
const int inf (0x3f3f3f3f);
template<typename T>void read(T &res){bool flag=false;char ch;while(!isdigit(ch=getchar()))(ch=='-')&&(flag=true);
for(res=ch-48;isdigit(ch=getchar());res=(res<<1)+(res<<3)+ch - 48);flag&&(res=-res);}
template<typename T>void Out(T x){if(x<0)putchar('-'),x=-x;if(x>9)Out(x/10);putchar(x%10+'0');}
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a*b/gcd(a,b);}
ll pow_mod(ll x,ll n,ll mod){ll res=1;while(n){if(n&1)res=res*x%mod;x=x*x%mod;n>>=1;}return res;}
ll fact_pow(ll n,ll p){ll res=0;while(n){n/=p;res+=n;}return res;}
ll mult(ll a,ll b,ll p){a%=p;b%=p;ll r=0,v=a;while(b){if(b&1){r+=v;if(r>p)r-=p;}v<<=1;if(v>p)v-=p;b>>=1;}return r;}
ll quick_pow(ll a,ll b,ll p){ll r=1,v=a%p;while(b){if(b&1)r=mult(r,v,p);v=mult(v,v,p);b>>=1;}return r;}
bool CH(ll a,ll n,ll x,ll t)
{ll r=quick_pow(a,x,n);ll z=r;for(ll i=1;i<=t;i++){r=mult(r,r,n);if(r==1&&z!=1&&z!=n-1)return true;z=r;}return r!=1;}
bool Miller_Rabin(ll n)
{if(n<2)return false;if(n==2)return true;if(!(n&1))return false;ll x=n-1,t=0;while(!(x&1)){x>>=1;t++;}srand(time(NULL));
ll o=8;for(ll i=0;i<o;i++){ll a=rand()%(n-1)+1;if(CH(a,n,x,t))return false;}return true;}
/*
int prime[30000010],minprime[30000010];
void euler(int n)
{int c=0,i,j;for(i=2;i<=n;i++){if(!minprime[i])prime[++c]=i,minprime[i]=i;for(j=1;j<=c&&i*prime[j]<=n;j++)
{minprime[i*prime[j]]=prime[j];if(i%prime[j]==0)break;}}}
struct Tree{ll l,r,sum,lazy,maxn,minn;}tree[1000000];
il void push_up(ll rt)
{tree[rt].sum=tree[rt<<1].sum+tree[rt<<1|1].sum;
tree[rt].maxn=max(tree[rt<<1].maxn,tree[rt<<1|1].maxn);
tree[rt].minn=min(tree[rt<<1].minn,tree[rt<<1|1].minn);}
il void push_down(ll rt , ll now)
{if(tree[rt].lazy){tree[rt<<1].lazy+=tree[rt].lazy;tree[rt<<1|1].lazy+=tree[rt].lazy;
tree[rt<<1].sum+=(now-(now>>1))*tree[rt].lazy;tree[rt<<1|1].sum+=(now>>1)*tree[rt].lazy;
tree[rt<<1].minn+=tree[rt].lazy;tree[rt<<1|1].minn+=tree[rt].lazy;
tree[rt<<1].maxn+=tree[rt].lazy;tree[rt<<1|1].maxn+=tree[rt].lazy;tree[rt].lazy=0;}}
il void build(ll l , ll r , ll rt , ll *aa)
{tree[rt].lazy=0;tree[rt].l=l;tree[rt].r=r;if(l==r)
{tree[rt].sum=aa[l];tree[rt].minn=tree[rt].sum;tree[rt].maxn=tree[rt].sum;return;}
ll mid=(l+r)>>1;build(l,mid,rt<<1,aa);build(mid+1,r,rt<<1|1,aa);push_up(rt);}
il void update_range(ll L , ll R , ll key , ll rt)
{if(L<=tree[rt].l&&R>=tree[rt].r){tree[rt].sum+=(tree[rt].r-tree[rt].l+1)*key;
tree[rt].minn+=key;tree[rt].maxn+=key;tree[rt].lazy+=key;return;}push_down(rt,tree[rt].r-tree[rt].l+1);
ll mid=(tree[rt].r+tree[rt].l)>>1;if(L<=mid)update_range(L,R,key,lson);
if(R>mid)update_range(L,R,key,rson);push_up(rt);}
il ll query_range(ll L, ll R, ll rt)
{if(L<=tree[rt].l&&R>=tree[rt].r){return tree[rt].sum;}push_down(rt,tree[rt].r-tree[rt].l+1);
ll mid=(tree[rt].r+tree[rt].l)>>1;ll ans=0;if(L<=mid)ans+=query_range(L,R,lson);
if(R>mid)ans+=query_range(L,R,rson);return ans;}
il ll query_min(ll L,ll R,ll rt)
{if(L<=tree[rt].l&&R>=tree[rt].r){return tree[rt].minn;}push_down(rt,tree[rt].r-tree[rt].l+1);
ll mid=(tree[rt].r+tree[rt].l)>>1;ll ans=(0x3f3f3f3f3f3f3f3fll);if(L<=mid)ans=min(ans,query_min(L,R,lson));
if(R>mid)ans=min(ans,query_min(L,R,rson));return ans;}
il ll query_max(ll L, ll R, ll rt)
{if(L<=tree[rt].l&&R>=tree[rt].r){return tree[rt].maxn;}push_down(rt,tree[rt].r-tree[rt].l+1);
ll mid=(tree[rt].r+tree[rt].l)>>1;ll ans=-(0x3f3f3f3f3f3f3f3fll);if(L<=mid)ans=max(ans,query_max(L,R,lson));
if(R>mid)ans=max(ans,query_max(L,R,rson));return ans;}
*/
const int N = 3e5 + 10;
ll a[N];
vector<ll>in , ou;
vector<pair<pair<ll , ll>, ll> > ans;
int main()
{
ios;
ll n , m ;
cin >> n >> m;
rep(i , 1 , m)
{
ll x , y , z;
cin >> x >> y >> z;
a[x] += z;
a[y] -= z;
}
rep(i , 1 , n)
{
if(a[i] > 0)
in.pb(i);
else if(a[i] < 0)
ou.pb(i);
}
ll one = 0 , two = 0;
while(true)
{
if(one == in.size() || two == ou.size())
break;
ll ha = min(a[in[one]] , -a[ou[two]]);
ans.pb(make_pair(make_pair(in[one] , ou[two]) , ha));
a[in[one]] -= ha;
a[ou[two]] += ha;
if(!a[in[one]]) one ++;
if(!a[ou[two]]) two ++;
}
ll len = ans.size();
cout << len << '\n';
rep(i , 0 , len - 1)
cout << ans[i].fi.fi << " " << ans[i].fi.se << " " << ans[i].se << "\n";
return 0;
}
View Code
E. Spaceship Solitaire
題目連結：https://codeforces.ml/contest/1266/problem/E
題意：
你要建個飛船 ， 建飛船需要 n 種資源 ， 每種資源的需求量為 a[i] 。每一回合你只能選擇一種資源並生產一個。然後你有里程牌？？？
里程碑包含三個元素 S 、 T 、 U ， 意思是如果你有 T 個 S 資源， 你就可以獲得一個U資源。（0 < T < a[S]）
不同的里程碑 S 和 T 不相同 （如果S 相同且 T 相同則它們是同一個里程碑）
給你 q 次詢問， 每次詢問會加入一個里程碑（如果該里程碑先前存在過 ， 則用現在的里程碑代替之前的里程碑 ， 如果U == 0 , 則銷燬該里程碑）
問建造飛船最少需要多少回合
分析：
i 種資源 每種需要 a[i] 個，那麼當沒有里程碑的情況下總的回合數為 a[1] + a[2] + ... + a[n] , 記為sum
首先我們要知道的是，不論如何，最後飛船是肯定可以建成的。又因為對任意里程碑 T < a[S] , 所以里程碑起作用的條件是肯定可以滿足的 ，那麼對於當前詢問加入的里程碑來說
如果它先前出現過，則我們判斷它先前出現的時候是否會對sum起影響（因為可能在該里程碑起作用之前，資源U已經採集的夠了，所以該里程碑不起到作用。如果在起作用前U採集還不夠，則它起到作用了， sum 會減一）
如果不影響，sum保持不變，影響了，則sum++（把先前造成的影響去掉）
然後我們判斷新的里程碑是否起作用（如果起作用sum -- ， 否則sum不變），對於每個詢問輸出sum即可。（只需判斷當前型別里程碑，其他里程碑不受影響）
簡單模擬題。
#include<bits/stdc++.h>
#define ios std::ios::sync_with_stdio(false)
#define sd(n) scanf("%d",&n)
#define sdd(n,m) scanf("%d%d",&n,&m)
#define sddd(n,m,k) scanf("%d%d%d",&n,&m,&k)
#define pd(n) printf("%d\n", (n))
#define pdd(n,m) printf("%d %d\n", n, m)
#define pld(n) printf("%lld\n", n)
#define pldd(n,m) printf("%lld %lld\n", n, m)
#define sld(n) scanf("%lld",&n)
#define sldd(n,m) scanf("%lld%lld",&n,&m)
#define slddd(n,m,k) scanf("%lld%lld%lld",&n,&m,&k)
#define sf(n) scanf("%lf",&n)
#define sff(n,m) scanf("%lf%lf",&n,&m)
#define sfff(n,m,k) scanf("%lf%lf%lf",&n,&m,&k)
#define rep(i,a,n) for (int i=a;i<=n;i++)
#define per(i,n,a) for (int i=n;i>=a;i--)
#define mm(a,n) memset(a, n, sizeof(a))
#define pb push_back
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define il inline
#define ll long long
#define lson rt << 1
#define rson rt << 1 | 1
#define MOD 1000000007
#define pi 3.14159265358979323
#define debug(x) cout <<#x<<": "<<x<<endl
#define debug2(x, y) cout <<#x<<": "<<x<<" | "<<#y<<": "<<y<<endl;
#define debug3(x, y, z) cout <<#x<<": "<<x<<" | "<<#y<<": "<<y<<" | "<<#z<<": "<<z<<endl;
#define debug4(a, b, c, d) cout <<#a<<": "<<a<<" | "<<#b<<": "<<b<<" | "<<#c<<": "<<c<<" | "<<#d<<": "<<d<<endl;
using namespace std;
const ll INF (0x3f3f3f3f3f3f3f3fll);
const int inf (0x3f3f3f3f);
template<typename T>void read(T &res){bool flag=false;char ch;while(!isdigit(ch=getchar()))(ch=='-')&&(flag=true);
for(res=ch-48;isdigit(ch=getchar());res=(res<<1)+(res<<3)+ch - 48);flag&&(res=-res);}
template<typename T>void Out(T x){if(x<0)putchar('-'),x=-x;if(x>9)Out(x/10);putchar(x%10+'0');}
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a*b/gcd(a,b);}
ll pow_mod(ll x,ll n,ll mod){ll res=1;while(n){if(n&1)res=res*x%mod;x=x*x%mod;n>>=1;}return res;}
ll fact_pow(ll n,ll p){ll res=0;while(n){n/=p;res+=n;}return res;}
ll mult(ll a,ll b,ll p){a%=p;b%=p;ll r=0,v=a;while(b){if(b&1){r+=v;if(r>p)r-=p;}v<<=1;if(v>p)v-=p;b>>=1;}return r;}
ll quick_pow(ll a,ll b,ll p){ll r=1,v=a%p;while(b){if(b&1)r=mult(r,v,p);v=mult(v,v,p);b>>=1;}return r;}
bool CH(ll a,ll n,ll x,ll t)
{ll r=quick_pow(a,x,n);ll z=r;for(ll i=1;i<=t;i++){r=mult(r,r,n);if(r==1&&z!=1&&z!=n-1)return true;z=r;}return r!=1;}
bool Miller_Rabin(ll n)
{if(n<2)return false;if(n==2)return true;if(!(n&1))return false;ll x=n-1,t=0;while(!(x&1)){x>>=1;t++;}srand(time(NULL));
ll o=8;for(ll i=0;i<o;i++){ll a=rand()%(n-1)+1;if(CH(a,n,x,t))return false;}return true;}
/*
int prime[30000010],minprime[30000010];
void euler(int n)
{int c=0,i,j;for(i=2;i<=n;i++){if(!minprime[i])prime[++c]=i,minprime[i]=i;for(j=1;j<=c&&i*prime[j]<=n;j++)
{minprime[i*prime[j]]=prime[j];if(i%prime[j]==0)break;}}}
struct Tree{ll l,r,sum,lazy,maxn,minn;}tree[1000000];
il void push_up(ll rt)
{tree[rt].sum=tree[rt<<1].sum+tree[rt<<1|1].sum;
tree[rt].maxn=max(tree[rt<<1].maxn,tree[rt<<1|1].maxn);
tree[rt].minn=min(tree[rt<<1].minn,tree[rt<<1|1].minn);}
il void push_down(ll rt , ll now)
{if(tree[rt].lazy){tree[rt<<1].lazy+=tree[rt].lazy;tree[rt<<1|1].lazy+=tree[rt].lazy;
tree[rt<<1].sum+=(now-(now>>1))*tree[rt].lazy;tree[rt<<1|1].sum+=(now>>1)*tree[rt].lazy;
tree[rt<<1].minn+=tree[rt].lazy;tree[rt<<1|1].minn+=tree[rt].lazy;
tree[rt<<1].maxn+=tree[rt].lazy;tree[rt<<1|1].maxn+=tree[rt].lazy;tree[rt].lazy=0;}}
il void build(ll l , ll r , ll rt , ll *aa)
{tree[rt].lazy=0;tree[rt].l=l;tree[rt].r=r;if(l==r)
{tree[rt].sum=aa[l];tree[rt].minn=tree[rt].sum;tree[rt].maxn=tree[rt].sum;return;}
ll mid=(l+r)>>1;build(l,mid,rt<<1,aa);build(mid+1,r,rt<<1|1,aa);push_up(rt);}
il void update_range(ll L , ll R , ll key , ll rt)
{if(L<=tree[rt].l&&R>=tree[rt].r){tree[rt].sum+=(tree[rt].r-tree[rt].l+1)*key;
tree[rt].minn+=key;tree[rt].maxn+=key;tree[rt].lazy+=key;return;}push_down(rt,tree[rt].r-tree[rt].l+1);
ll mid=(tree[rt].r+tree[rt].l)>>1;if(L<=mid)update_range(L,R,key,lson);
if(R>mid)update_range(L,R,key,rson);push_up(rt);}
il ll query_range(ll L, ll R, ll rt)
{if(L<=tree[rt].l&&R>=tree[rt].r){return tree[rt].sum;}push_down(rt,tree[rt].r-tree[rt].l+1);
ll mid=(tree[rt].r+tree[rt].l)>>1;ll ans=0;if(L<=mid)ans+=query_range(L,R,lson);
if(R>mid)ans+=query_range(L,R,rson);return ans;}
il ll query_min(ll L,ll R,ll rt)
{if(L<=tree[rt].l&&R>=tree[rt].r){return tree[rt].minn;}push_down(rt,tree[rt].r-tree[rt].l+1);
ll mid=(tree[rt].r+tree[rt].l)>>1;ll ans=(0x3f3f3f3f3f3f3f3fll);if(L<=mid)ans=min(ans,query_min(L,R,lson));
if(R>mid)ans=min(ans,query_min(L,R,rson));return ans;}
il ll query_max(ll L, ll R, ll rt)
{if(L<=tree[rt].l&&R>=tree[rt].r){return tree[rt].maxn;}push_down(rt,tree[rt].r-tree[rt].l+1);
ll mid=(tree[rt].r+tree[rt].l)>>1;ll ans=-(0x3f3f3f3f3f3f3f3fll);if(L<=mid)ans=max(ans,query_max(L,R,lson));
if(R>mid)ans=max(ans,query_max(L,R,rson));return ans;}
*/
const int N = 2e5 + 10;
ll a[N] , sum;
map<ll , ll> vis[N];
int main()
{
ios;cin.tie(0);
int n , q;
cin >> n ;
rep(i , 1 , n)
cin >> a[i] , sum += a[i];
cin >> q;
rep(i , 1 , q)
{
ll s , t , u;
cin >> s >> t >> u;
if(vis[s][t])
{
a[vis[s][t]] ++;
if(a[vis[s][t]] > 0)
sum ++;
vis[s][t] = 0;
}
if(u)
{
vis[s][t] = u;
a[u] --;
if(a[u] >= 0)
sum --;
}
cout << sum << '\n';
}
return 0;
}
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