第十五屆浙江大學寧波理工學院程式設計大賽
阿新 • • 發佈：2018-12-15
A題：簽到題 1998+n
#include<stdio.h> #include<iostream> #include<algorithm> #include<cmath> #include<cstring> #include<string.h> #include<queue> #include<stack> #include<list> #include<map> #include<set> #include<vector> using namespace std; typedef long long int ll; const int maxn =100+5; const int maxm=10000; const int mod =1e9+7; const int INF=0x3f3f3f3f; const double eps=1e-8; int main() { int n;scanf("%d",&n); printf("%d\n",1998+n); return 0; }
B題：
n+1
證明略
#include<stdio.h> #include<iostream> #include<algorithm> #include<cmath> #include<cstring> #include<string.h> #include<queue> #include<stack> #include<list> #include<map> #include<set> #include<vector> using namespace std; typedef long long int ll; const int maxn =100+5; const int maxm=10000; const int mod =1e9+7; const int INF=0x3f3f3f3f; const double eps=1e-8; int main() { int t;scanf("%d",&t); while(t--) { int n;scanf("%d",&n); printf("%d\n",n+1); } return 0; }
D
二進位制列舉答案，然後維護一個可行的l,r，當可行的l,r包含了輸入的個數的時候，就是滿足條件的，然後從所有答案中選出最大的就行。
#include<stdio.h> #include<iostream> #include<algorithm> #include<cmath> #include<cstring> #include<string.h> #include<queue> #include<stack> #include<list> #include<map> #include<set> #include<vector> using namespace std; typedef long long int ll; typedef unsigned long long ull; const int maxn =20010; const int maxm=10000; const int mod =998244353; const int INF=0x3f3f3f3f; const double eps=1e-8; int l[10],r[10]; int main() { int T;scanf("%d",&T); while(T--) { int n,ans=0; scanf("%d",&n); for(int i=0;i<7;i++) scanf("%d%d",&l[i],&r[i]); for(int j=0;j<(1<<7);j++) { long long mi=0,mx=0;int tot=0; for(int i=0;i<7;i++) { if((j>>i)&1) mi+=l[i],mx+=r[i],++tot; } if(n>=mi&&n<=mx) ans=std::max(ans,tot); } printf("%d\n",ans); } }
H
快速冪
#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string.h>
#include<queue>
#include<stack>
#include<list>
#include<map>
#include<set>
#include<vector>
using namespace std;
typedef long long int ll;
typedef unsigned long long ull;
const int maxn =20010;
const int maxm=10000;
const int mod =998244353;
const int INF=0x3f3f3f3f;
const double eps=1e-8;
ll fpow(ll a,ll b)
{
ll ans=1;
while(b)
{
if(b&1)ans=ans*a%mod;
a=a*a%mod;
b>>=1;
}
return ans;
}
int main()
{
int t;scanf("%d",&t);
while(t--)
{
ll ans=1;
int n;scanf("%d",&n);
while(n--)
{
ll a,b;scanf("%lld%lld",&a,&b);
ans*=(fpow(2,a)-1+fpow(2,b)-1)%mod;
ans%=mod;
}
printf("%lld\n",ans);
}
return 0;
}I
模擬就行了
#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string.h>
#include<queue>
#include<stack>
#include<list>
#include<map>
#include<set>
#include<vector>
using namespace std;
typedef long long int ll;
typedef unsigned long long ull;
const int maxn =20010;
const int maxm=10000;
const int mod =998244353;
const int INF=0x3f3f3f3f;
const double eps=1e-8;
int main(){
char f[1005];
int T,t[1005];
scanf("%d",&T);
while(T--)
{
int n;
scanf("%d",&n);
for(int i=0;i<n;i++)
{
char race[30];
scanf("%d is %s",&t[i],race);
f[i]=race[0];
}
char s[30];
scanf("%s",s);
int len=strlen(s);
for(int i=0;i<len;i++)
{
if(s[i]=='T'||s[i]=='Z'||s[i]=='P')
{
for(int j=0;j<n;j++)
{
if(f[j]==s[i]) printf("%d ",t[j]);
}
}
}
printf("\n");
}
return 0;
}J
#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string.h>
#include<queue>
#include<stack>
#include<list>
#include<map>
#include<set>
#include<vector>
using namespace std;
typedef long long int ll;
typedef unsigned long long ull;
const int maxn =1e5+5;
const int maxm=10000;
const int mod =1e9+7;
const int INF=0x3f3f3f3f;
const double eps=1e-8;
int main()
{
int t;scanf("%d",&t);
while(t--)
{
int a,b,x,y;scanf("%d%d%d%d",&a,&b,&x,&y);
if(a<350)
{
printf("You have not enough minerals.\n");
continue;
}
if(b<250)
{
printf("You require more vespene gas.\n");
continue;
}
if(x+6>y)
{
printf("You must construct additional pylons.\n");
continue;
}
printf("Carrier has arrived.\n");
}
return 0;
}K
模擬
#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string.h>
#include<queue>
#include<stack>
#include<list>
#include<map>
#include<set>
#include<vector>
using namespace std;
typedef long long int ll;
typedef unsigned long long ull;
const int maxn =20010;
const int maxm=10000;
const int mod =1e9+7;
const int INF=0x3f3f3f3f;
const double eps=1e-8;
int a[maxn];
int b[maxn];
int main()
{
int t;scanf("%d",&t);
while(t--)
{
int n,q;scanf("%d%d",&n,&q);
for(int i=1;i<=n;i++) scanf("%d%d",&a[i],&b[i]);
for(int i=2;i<=n;i++)
{
int f;scanf("%d",&f);
a[i]+=a[f];
b[i]+=b[f];
}
while(q--)
{
int x;scanf("%d",&x);
printf("%d %d\n",a[x],b[x]);
}
}
return 0;
}