A. Apple

按題意模擬即可。

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<string>
#include<ctype.h>
#include<math.h>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<bitset>
#include<algorithm>
#include<time.h>
using namespace std;
void fre() { freopen("c://test//input.in", "r", stdin); freopen("c://test//output.out", "w", stdout); }
#define MS(x, y) memset(x, y, sizeof(x))
#define ls o<<1
#define rs o<<1|1
typedef long long LL;
typedef unsigned long long UL;
typedef unsigned int UI;
template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b > a)a = b; }
template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b < a)a = b; }
const int N = 105, M = 0, Z = 1e9 + 7, inf = 0x3f3f3f3f;
template <class T1, class T2>inline void gadd(T1 &a, T2 b) { a = (a + b) % Z; }
int casenum, casei;
int n, m;
vector<int>a[N];

int ind1 = 0, ind1num = 0;
int ind2num = 0;
int ind3 = 0, ind3num = 0;
bool vis[N];
int ind[N];

int cnt;
bool dfs1(int x, int fa)
{
    ++cnt;
    vis[x] = 1;
    for(auto y : a[x])if(y != fa)
    {
        if(!vis[y])
        {
            if(!dfs1(y, x))return 0;
        }
        else if(y != ind3)return 0;
    }
    return 1;
}

void dfs3(int x, int fa)
{
    ++cnt;
    vis[x] = 1;
    for(auto y : a[x])if(y != fa)
    {
        if(!vis[y])
        {
            dfs3(y, x);
        }
    }
}

bool solve()
{
    if(ind1num != 1)return 0;
    if(ind2num != n - 2)return 0;
    if(ind3num != 1)return 0;

    cnt = 0;
    vis[ind3] = true;
    if(!dfs1(ind1, 0))return 0;
    dfs3(ind3, 0);
    if(cnt != n)return 0;

    return 1;
}
int main()
{
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
	while(~scanf("%d%d", &n, &m))
	{
        for(int i = 1; i <= n; ++i)
        {
            a[i].clear();
            ind[i] = 0;
            vis[i] = 0;
        }
        for(int i = 1; i <= m; ++i)
        {
            int x, y; scanf("%d%d", &x, &y);
            a[x].push_back(y);
            a[y].push_back(x);
            ++ind[x];
            ++ind[y];
        }
        ind1 = 0, ind1num = 0;
        ind2num = 0;
        ind3 = 0, ind3num = 0;
        for(int i = 1; i <= n; ++i)
        {
            if(ind[i] == 1)
            {
                ++ind1num;
                ind1 = i;
            }
            else if(ind[i] == 2)
            {
                ++ind2num;
            }
            else if(ind[i] == 3)
            {
                ++ind3num;
                ind3 = i;
            }
        }
        puts(solve() ? "Yes" : "No");
	}

	return 0;
}
/*
【trick&&吐槽】


【題意】


【分析】


【時間復雜度&&優化】
9 9
1 2
2 3
3 5
5 6
6 4
4 7
1 7
7 9
8 9

5 5
1 2
2 3
1 3
1 4
1 5

4 4
1 2
2 3
1 3
1 4

5 4
1 2
2 3
1 3
1 4


*/

B. Bar charts

關於序列的前綴和建立差分約束系統，SPFA判斷是否存在負環。

#include<cstdio>
#include<cstdlib>
using namespace std;
const int N=12222,M=1000000,E=10000000,inf=~0U>>1;
int n,i,g[N],v[M],w[M],nxt[M],ed,d[N];
int x,h,t,q[E];
int vis[N],in[N];
inline void add(int x,int y,int z){v[++ed]=y;w[ed]=z;nxt[ed]=g[x];g[x]=ed;}
inline void make(int x,int y,int z){
    //x-y=z
    add(y,x,z);
    add(x,y,-z);
}
void gao(){
    int k,b,s;
    scanf("%d%d%d",&k,&b,&s);
    make(b-1,0,0);
    for(int i=b+k*s;i<=n;i++)make(i,i-1,0);
    for(int i=1;i<=k;i++){
        int l=b+(i-1)*s,r=b+i*s;
        int o;
        scanf("%d",&o);
        //s[r-1]-s[l-1]
        make(r-1,l-1,o);
    }
}
void NO(){
    puts("No");
    exit(0);
}
inline void ext(int x,int y){
    if(y<0)NO();
    if(y>=d[x])return;
    d[x]=y;
    if(!in[x]){
        in[x]=1;
        vis[x]++;
        if(vis[x]>=n)NO();
        q[++t]=x;
        if(t>=E-10)NO();
    }
}
int main(){
    freopen("input.txt","r",stdin);
    freopen("output.txt","w",stdout);
    n=11111;
    for(i=0;i<n;i++)add(i+1,i,0);
    gao();
    gao();
    for(i=0;i<=n;i++){
        d[i]=inf;
    }
    h=1,t=0;
    ext(0,0);
    while(h<=t){
        x=q[h++];
        for(i=g[x];i;i=nxt[i])ext(v[i],d[x]+w[i]);
        in[x]=0;
    }
    puts("Yes");
}

C. Construction sets

二分答案，二進制拆分背包+bitset檢驗。

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<string>
#include<ctype.h>
#include<math.h>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<bitset>
#include<algorithm>
#include<time.h>
using namespace std;
void fre() { freopen("c://test//input.in", "r", stdin); freopen("c://test//output.out", "w", stdout); }
#define MS(x, y) memset(x, y, sizeof(x))
#define ls o<<1
#define rs o<<1|1
typedef long long LL;
typedef unsigned long long UL;
typedef unsigned int UI;
template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b > a)a = b; }
template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b < a)a = b; }
const int N = 55, M = 0, Z = 1e9 + 7, inf = 0x3f3f3f3f;
template <class T1, class T2>inline void gadd(T1 &a, T2 b) { a = (a + b) % Z; }
int casenum, casei;
int n, mn, mx;
int m[N], c[N];
bitset<10005>f;
bool check(int K)
{
    f.reset(); f[0] = 1;
    for(int i = 1; i <= n; ++i)
    {
        int tot = c[i] / K;
        int g = 1;
        LL v = m[i];
        while(tot)
        {
            if(v > mx)break;
            f |= f << v;
            tot -= g;
            g = min(g * 2, tot);
            v = (LL)m[i] * g;
        }
    }
    for(int j = mn; j <= mx; ++j)
    {
        if(f[j])return 1;
    }
    return 0;
}
int main()
{
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
	while(~scanf("%d%d%d", &n, &mn, &mx))
	{
        for(int i = 1; i <= n; ++i)
        {
            scanf("%d%d", &m[i], &c[i]);
        }
        int l = 0;
        int r = 1e6;
        while(l < r)
        {
            int mid = (l + r + 1) >> 1;
            if(check(mid))
            {
                l = mid;
            }
            else
            {
                r = mid - 1;
            }
        }
        printf("%d\n", l);
	}

	return 0;
}
/*
【trick&&吐槽】


【題意】


【分析】


【時間復雜度&&優化】
3 12 13
3 8
4 6
7 9

*/

D. Dinner party

$f[i][j]$表示面積和為$i$的矩形，周長和為$j$是否有可能，bitset加速。

#include<cstdio>
#include<bitset>
using namespace std;
const int N=1010,M=2010;
int T,n,m,i,j,x,y,cnt;
bitset<M>f[N];
void solve(){
    scanf("%d%d",&n,&m);
    if(m%2){
        puts("No");
        return;
    }
    m/=2;
    if(f[n][m]==0){
        puts("No");
        return;
    }
    puts("Yes");
    int cnt=0;
    static int q[100000][2];
    while(n){
        int X=0,Y=0;
        for(x=1;x<=n;x++){
            for(y=1;y*x<=n;y++)if(x+y<=m)if(f[n-x*y][m-x-y]){
                X=x,Y=y;
                break;
            }
            if(X)break;
        }
        q[++cnt][0]=X;
        q[cnt][1]=Y;
        n-=X*Y;
        m-=X+Y;
    }
    printf("%d\n",cnt);
    for(int i=1;i<=cnt;i++)printf("%d %d\n",q[i][0],q[i][1]);
}
int main(){
    freopen("input.txt","r",stdin);
    freopen("output.txt","w",stdout);
    //scanf("%d%d",&n,&m);
    n=1000;
    m=2000;
    f[0][0]=1;
    for(i=0;i<=n;i++)
        for(x=1;x+i<=n;x++)
            for(y=1;x*y+i<=n;y++)
                f[i+x*y]|=f[i]<<(x+y);
    scanf("%d",&T);
    while(T--)solve();
}

E. Expression on dice

若要構造$=$，則在兩側構造$0$即可，否則隨意構造。

//
//  main.cpp
//  opencup 10378 E
//
//  Created by luras on 2017/9/14.
//  Copyright © 2017年 luras. All rights reserved.
//

#define ms(x, y) memset(x, y, sizeof(x))
#define mc(x, y) memcpy(x, y, sizeof(x))
#define lson o << 1, l, mid
#define rson o << 1 | 1, mid + 1, r
#define ls o << 1
#define rs o << 1 | 1
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<queue>
#include<map>
#include<stack>
#include<time.h>
#include<vector>
#include<list>
#include<set>
#include<iostream>
#include<stdlib.h>
#include<string>
#include<algorithm>
#pragma comment(linker,"/STACK:102400000,102400000")
template <class T> inline void gmax(T &a, T b){if(b > a) a = b;}
template <class T> inline void gmin(T &a, T b){if(b < a) a = b;}
using namespace std;
const int N = 1e6 + 10, M = 2e6 + 10, Z = 1e9 + 7, maxint = 2147483647, ms1 = 16843009, ms31 = 522133279, ms63 = 1061109567, ms127 = 2139062143;
const double PI = acos(-1.0), eps = 1e-8;
typedef long long LL;
void fre()
{
    freopen("/Users/luras/Desktop/in.txt", "r", stdin);
    freopen("/Users/luras/Desktop/out.txt", "w", stdout);
}
const int INF = 1e9;
int casenum, casei;

int priv[N];
double calc(double a, double b, char op)
{
    if(op == ‘+‘) return a + b;
    if(op == ‘-‘) return a - b;
    if(op == ‘*‘) return a * b;
    if(op == ‘/‘) return a / b;
    return 0;
}


double calculate(string str)
{
    stack<double> num;
    stack<char> oper;
    priv[‘+‘] = priv[‘-‘] = 3;
    priv[‘*‘] = priv[‘/‘] = 2;
    priv[‘(‘] = 10;
    double x, y, t = 0;
    int i; char last = 0;
    for(i = 0; i < str.length(); i ++){
        if(isdigit(str[i])){
            num.push(atof(str.c_str() + i));
            for(; i + 1 < str.length() && isdigit(str[i + 1]); i ++);
            if(i + 1 < str.length() && str[i + 1] == ‘.‘)
                for(i ++; i + 1 < str.length() && isdigit(str[i + 1]); i ++);
        }
        else if(str[i] == ‘(‘){
            oper.push(str[i]);
        }
        else if(str[i] == ‘)‘){
            while(oper.top() != ‘(‘){
                y = num.top(); num.pop();
                x = num.top(); num.pop();
                char op = oper.top();
                oper.pop();
                num.push(calc(x, y, op));
            }
            oper.pop();
        }
        else if(str[i] == ‘-‘ && (last == 0 || last == ‘(‘)){
            num.push(0.0);
            oper.push(‘-‘);
        }
        else if(priv[str[i]] > 0){
            while(oper.size() > 0 && priv[str[i]] >= priv[oper.top()]){
                y = num.top(); num.pop();
                x = num.top(); num.pop();
                char op = oper.top();
                oper.pop();
                num.push(calc(x, y, op));
            }
            oper.push(str[i]);
        }else continue;
        last = str[i];
    }
    while(oper.size() > 0){
        y = num.top(); num.pop();
        x = num.top(); num.pop();
        char op = oper.top();
        oper.pop();
        num.push(calc(x, y, op));
    }
    return num.top();
}

map<int, bool> mop;
vector<int> num;
vector<char> sym;
int lft, rgt, zero, divi, typ;


const string sta[6][6] ={
    {"=", "<", ">", "!=", "<=", ">="},
    {"4", "+", "-", "(", "(", ")"},
    {"0", "/", "/", "/", "8", "+"},
    {"2", "3", "4", "5", "-", ")"},
    {"+", "-", "*", "/", "1", "9"},
    {"6", "7", "+", "-", "(", ")"}
};


void ask(int o)
{
    char ch;
    if(o == 2){
        puts("2");
        fflush(stdout);
        scanf(" %c", &ch);
        //ch = sta[1][rand() % 6][0];
        //cout << "A: " << ch;
        if(ch == ‘(‘) lft ++;
        else if(ch == ‘)‘) rgt ++;
        else if(ch == ‘4‘) {
            num.push_back(ch);
            if(mop[ch] == 0){
                mop[ch] = 1;
                typ ++;
            }
        }
        else sym.push_back(ch);
    }
    else if(o == 3){
        puts("3");
        fflush(stdout);
        scanf(" %c", &ch);
        //ch = sta[2][rand() % 6][0];
        //cout << "A: " << ch;
        
        if(ch == ‘0‘) zero ++;
        else if(ch == ‘8‘) {
            num.push_back(ch);
            if(mop[ch] == 0){
                mop[ch] = 1;
                typ ++;
            }
        }
        else if(ch == ‘/‘) divi ++;
        else sym.push_back(ch);
    }
    else if(o == 4){
        puts("4");
        fflush(stdout);
        //ch = sta[3][rand() % 6][0];
        //cout << "A: " << ch;
        
        scanf(" %c", &ch);
        if(ch == ‘-‘) sym.push_back(ch);
        else if(ch == ‘)‘) rgt ++;
        else {
            num.push_back(ch);
            if(mop[ch] == 0){
                mop[ch] = 1;
                typ ++;
            }
        }
    }
}


string ans;
string s;
string ss;

int sgn(double x)
{
    if(fabs(x) < eps) return 0;
    return x > 0 ? 1 : -1;
}

bool choose()
{
    s = "";
    int DIV = divi, ZERO = zero;
    int n = num.size(), m = sym.size() + DIV;
    int bas = (n + ZERO) / (m + 1);
    int mo = (n + ZERO) - (m + 1) * bas;
    int t = 0, j = -1, k = -1;
    for(int i = 0; i < m + 1; i ++){
        if(mo){
            if(j + 1 >= n) return 0; // 必須得放前導0了
            s += num[++ j]; ++ t;
            while(t <= bas && ZERO){
                s += ‘0‘; t ++; ZERO --;
            }
            while(t <= bas){
                s += num[++ j]; t ++;
            }
            mo --;
            t = 0;
        }
        else{
            if(j + 1 >= n) return 0;
            s += num[++ j]; ++ t;
            while(t < bas && ZERO){
                s += ‘0‘; t ++; ZERO --;
            }
            while(t < bas){
                s += num[++ j]; t ++;
            }
            t = 0;
        }
        if(DIV){
            s += ‘/‘;
            DIV --;
        }
        else{
            if(k + 1 < sym.size())s += sym[++ k];
        }
    }
    return 1;
}

bool check(int o)
{
    if(o == 0){
        zero -= 2; lft --; rgt --; divi --;
        if(!choose()) {zero += 2; lft ++; rgt ++; divi ++; return 0;}
        if(calculate(s) != 0){
            ans = "";
            for(int i = 0; i < lft; i ++) ans += ‘(‘;
            ans += ‘0‘;
            for(int i = 0; i < rgt; i ++) ans += ‘)‘;
            ans += "=0/(" + s + ‘)‘;
        }
        else {zero += 2; lft ++; rgt ++; divi ++; return 0;}
    }
    else{
        zero --;
        if(!choose()) {zero ++;return 0;}
        double tmp = calculate(s);
        if(sgn(tmp) != 0){
            if(ss[0] == ‘<‘ && sgn(tmp) > 0 || ss[0] == ‘>‘ && sgn(tmp) < 0){
                ans = "";
                for(int i = 0; i < lft; i ++) ans += ‘(‘;
                ans += ‘0‘;
                for(int i = 0; i < lft; i ++) ans += ‘)‘;
                ans += ss + s;
            }
            else{
                ans = "";
                for(int i = 0; i < lft; i ++) ans += ‘(‘;
                ans += s;
                for(int i = 0; i < lft; i ++) ans += ‘)‘;
                ans += ss + ‘0‘;
            }
        }
        else if(ss == "!="){
            ans = "";
            for(int i = 0; i < lft; i ++) ans += ‘(‘;
            ans += ‘0‘;
            for(int i = 0; i < lft; i ++) ans += ‘)‘;
            ans += ss + s;
            
        }
        else {zero ++; return 0;}
    }
    return 1;
}

int main()
{
    srand(time(NULL));
    //fre();
    puts("1");
    fflush(stdout);
    cin >> ss;
    //ss = sta[0][rand() % 6];
    //cout << ss << endl;
    lft = rgt = zero = divi = typ = 0;
    int tim = 0;
    while(1){
        if(++ tim == 1000){
            int go = 1;
        }
        if(zero < (ss[0] == ‘=‘ ? 2 : 1) || divi == 0){
            ask(3);
        }
        else if(lft < rgt || lft == 0){
            ask(2);
        }
        else if(num.size() - 2 < sym.size() || rgt < lft){
            ask(4);
        }
        else {
            if(check(ss[0] == ‘=‘ ? 0 : 1)) break;
            ask(4);
        }
    }
    printf("0 ");
    cout << ans << endl;
    fflush(stdout);
    return 0;
}

/*
 
 
 題意:
 
 類型:
 
 分析:
 
 優化:
 
 trick:
 
 數據:
 
 Sample Input
 
 Sample Output
 
 >=
 + / / + 8 0 ) ( ) ( - - 5 5 5 - 4 4 4 - 5 3
 
 */

F. Flight trip

留坑。

G. Glasses with solutions

找出的子集需要滿足$b\sum m-a\sum t=0$，折半搜索即可。

#include<cstdio>
#include<map>
using namespace std;
typedef long long ll;
const int N=40;
int n,m,A,B,i,x,y,a[N];map<ll,ll>f;
ll ans;
void dfsl(int x,ll y){
    if(x==m){f[y]++;return;}
    dfsl(x+1,y+a[x]);
    dfsl(x+1,y);
}
void dfsr(int x,ll y){
    if(x==n){ans+=f[-y];return;}
    dfsr(x+1,y+a[x]);
    dfsr(x+1,y);
}
int main(){
    freopen("input.txt","r",stdin);
    freopen("output.txt","w",stdout);
    scanf("%d%d%d",&n,&A,&B);
    for(i=0;i<n;i++){
        scanf("%d%d",&x,&y);
        a[i]=x*B-A*y;
    }
    m=n/2;
    dfsl(0,0);
    dfsr(m,0);
    ans--;
    printf("%lld",ans);
}

H. Hamburgers

對於每個漢堡，$O(2^6)$枚舉所有它可以滿足的口味，然後對於每個人判斷是否滿足即可。

時間復雜度$O(m\sum a+2^6\sum b)$。

#include<cstdio>
#include<cstring>
#include<vector>
using namespace std;
const int N=55555;
int n,m,i,j,k,x,ans[N],mx[N],b[N];
vector<int>a[N];
bool v[(1<<26)+5];
inline int get(){
    static char s[1000];
    scanf("%s",s);
    int t=0,len=strlen(s);
    for(int i=0;i<len;i++)t|=1<<(s[i]-‘a‘);
    return t;
}
inline void make(int x,int y){
    for(int i=x;i;i=(i-1)&x)v[i]=y;
}
int main(){
    freopen("input.txt","r",stdin);
    freopen("output.txt","w",stdout);
    scanf("%d",&n);
    for(i=1;i<=n;i++){
        scanf("%d",&k);
        while(k--){
            x=get();
            a[i].push_back(x);
        }
        ans[i]=1;
    }
    scanf("%d",&m);
    for(i=1;i<=m;i++){
        scanf("%d",&k);
        for(j=1;j<=k;j++)b[j]=get();
        for(j=1;j<=k;j++)make(b[j],1);
        for(j=1;j<=n;j++){
            int t=0;
            for(x=0;x<a[j].size();x++)if(v[a[j][x]])t++;
            if(t>mx[j])mx[j]=t,ans[j]=i;
        }
        for(j=1;j<=k;j++)make(b[j],0);
    }
    for(i=1;i<=n;i++)printf("%d\n",ans[i]);
}
/*
5
1
a
3
a
ba
cb
3
vba
d
ba
3
ca
da
da
3
as
ba
af
2
4
abc
abcd
a
a
2
abcdef
abcdev
*/

I. Intricate path

留坑。

J. Jumps through the Hyperspace

顯然到每個點的時間越早越好，對於每種傳送門預處理出每種余數下的最優等待時間即可。

#include<cstdio>
#include<algorithm>
#include<vector>
#include<queue>
using namespace std;
typedef pair<int,int>P;
const int N=2010,inf=~0U>>1;
int n,m,st,i,j,a[N],b[N],c[N],d[N],w[N][N];
int f[N];
char g[N][N],op[N];
priority_queue<P,vector<P>,greater<P> >q;
inline void ext(int x,int y){
    if(f[x]>y)q.push(P(f[x]=y,x));
}
int main(){
    freopen("input.txt","r",stdin);
    freopen("output.txt","w",stdout);
    scanf("%d%d%d",&n,&m,&st);
    while(m--){
        scanf("%s",op);
        int o=op[0];
        scanf("%d%d%d%d",&a[o],&b[o],&c[o],&d[o]);
        for(i=0;i<c[o];i++){
            w[o][i]=inf;
            for(j=0;j<c[o];j++)w[o][i]=min(w[o][i],(a[o]*(i+j)+b[o])%c[o]+d[o]+j);
        }
    }
    for(i=1;i<=n;i++)scanf("%s",g[i]+1);
    for(i=1;i<=n;i++)f[i]=inf;
    ext(1,st);
    while(!q.empty()){
        P t=q.top();q.pop();
        if(f[t.second]<t.first)continue;
        for(i=1;i<=n;i++)if(g[t.second][i]!=‘.‘)
            ext(i,t.first+w[g[t.second][i]][t.first%c[g[t.second][i]]]);
    }
    if(f[n]==inf)puts("-1");else 
    printf("%d",f[n]-st);
}

K. King’s island

暴力搜索出一個上凸殼，然後翻轉拼接起來得到完整的多邊形即可。

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<string>
#include<ctype.h>
#include<math.h>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<bitset>
#include<algorithm>
#include<time.h>
using namespace std;
void fre() { freopen("c://test//input.in", "r", stdin); freopen("c://test//output.out", "w", stdout); }
#define MS(x, y) memset(x, y, sizeof(x))
#define ls o<<1
#define rs o<<1|1
typedef long long LL;
typedef unsigned long long UL;
typedef unsigned int UI;
template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b > a)a = b; }
template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b < a)a = b; }
const int N = 1010, M = 0, Z = 1e9 + 7, inf = 0x3f3f3f3f;
template <class T1, class T2>inline void gadd(T1 &a, T2 b) { a = (a + b) % Z; }
int casenum, casei;
int n;
struct B
{
    int x, y;
    bool operator < (const B & b)const
    {
        //y/x > b.y/b.x
        return y * b.x > x * b.y;
    }
};
B p[100000]; int g;
int f[N][N][62];
struct A
{
    int x, y, z;
};
bool bingo[60];
vector<B>ans[60];

void getvt(int x, int y, int v, vector<B>&vt, int DX, int DY)
{
    if(v == 0)return;
    int o = f[x][y][v];
    if(f[x - p[o].x][y - p[o].y][v] == o)
    {
        getvt(x - p[o].x, y - p[o].y, v, vt, DX + p[o].x, DY + p[o].y);
    }
    else
    {
        vt.push_back({p[o].x + DX, p[o].y + DY});
        getvt(x - p[o].x, y - p[o].y, v - 1, vt, 0, 0);
    }
}

void init()
{
    for(int i = 1; i <= 200; ++i)
    {
        for(int j = i + 1; j <= 200; ++j)if(__gcd(i, j) == 1)
        {
            int W = (i * i + j * j);
            int w = sqrt(W);
            if(w * w == W)
            {
                p[++g] = {i, j};
                p[++g] = {j, i};
            }
        }
    }
    sort(p + 1, p + g + 1);
    //printf("%d\n", g);

    MS(f, 0);
    f[0][0][0] = 1;
    for(int i = 0; i <= 500; ++i)
    {
        for(int j = 0; j <= 500; ++j)
        {
            for(int v = 29; v >= 0; --v)if(f[i][j][v])
            {
                //
                int k = f[i][j][v];
                auto &it = f[i + p[k].x][j + p[k].y][v];
                if(it == 0)it = k;
                //
                for(int k = f[i][j][v] + 1; k <= g; ++k)
                {
                    auto &it = f[i + p[k].x][j + p[k].y][v + 1];
                    if(it == 0)it = k;
                }
                if(i == 0 || j == 0)continue;
                for(int u = 1; u + v <= 30; ++u)if(f[i][j][u] && f[i][j][u] != f[i][j][v])
                {
                    if(!bingo[u + v])
                    {
                        bingo[u + v] = 1;
                        vector<B>up, down;
                        getvt(i, j, u, up, 0, 0);
                        getvt(i, j, v, down, 0, 0);
                        int x = i;
                        int y = j;
                        //printf("%d %d %d\n", u + v, up.size(), down.size());
                        //puts("up------------");
                        for(int r = 0; r < up.size(); ++r)
                        {
                            x -= up[r].x;
                            y -= up[r].y;
                            ans[u + v].push_back({x, y});
                            //printf("%d %d\n", up[r].x, up[r].y);
                        }
                        //puts("down------------");
                        for(int r = 0; r < down.size(); ++r)
                        {
                            x += down[r].x;
                            y += down[r].y;
                            ans[u + v].push_back({x, y});
                            //printf("%d %d\n", down[r].x, down[r].y);
                        }
                        //puts("ans------");
                        for(auto w : ans[u + v])
                        {
                            //printf("%d %d\n", w.x, w.y);
                        }
                        int pasue = 1;
                    }
                }
            }
        }
    }
    for(int i = 1;i <= 30; ++i)
    {
        //printf("%d\n", bingo[i]);
    }
}
int main()
{
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
    init();
	while(~scanf("%d", &n))
	{
	    if(n == 3)
        {
            puts("0 0\n4 3\n-20 21");
            continue;
        }
        for(auto it : ans[n])
        {
            printf("%d %d\n", it.x, it.y);
        }
	}

	return 0;
}
/*
【trick&&吐槽】


【題意】


【分析】


【時間復雜度&&優化】
28 195
9 15
-116 178
4
20 21
1 -159
-236 -11
-92 6
5
20 50
3 -94
-16 -274
-236 18
-92 35


*/

L. Lexica

無視同一行/同一列的兩個格子的位置關系，只需要在它們不相同時將方案數乘$2$，如此一來只關心每一行/每一列有多少格子沒滿足。

設$f[i][j][S][k]$表示考慮到$(i,j)$，每一列剩余情況為$S$，第$i$行還有$k$個格子需要填充時的方案數，然後逐格轉移即可。

時間復雜度$O(n3^n)$。

#include<cstdio>
typedef unsigned long long ll;
const int N=18,M=1600000;
int n,m,all,i,j,k,x,y,z,A,B,C,S,o,p[N],mask,cnt[N];
char a[N][N];
int e;
ll f[2][M][3],mul;
int w[M];
inline void clr(){
    for(int i=0;i<=mask;i++)for(int j=0;j<3;j++)f[e^1][i][j]=0;
}
inline void nxt(){
    //for(int i=0;i<=mask;i++)for(int j=0;j<3;j++)f[i][j]=g[i][j];
}
inline int get(int S,int x){
    return S/p[x]%3;
}
void dfs(int x,int y,int z){
		if(x==n){
			w[y]=z;
			return;
		}
		for(int i=0;i<3;i++)dfs(x+1,y*3+i,z*2+(!!i));
}
int main(){
    freopen("input.txt","r",stdin);
    freopen("output.txt","w",stdout);
    scanf("%d",&n);
    m=n+2;
    for(i=0;i<m;i++)scanf("%s",a[i]);
    mul=1;
    for(p[0]=i=1;i<N;i++)p[i]=p[i-1]*3;
    all=p[n];
    for(i=1;i<=n;i++){
        if(a[0][i]!=‘#‘&&a[m-1][i]!=‘#‘&&a[0][i]!=a[m-1][i])mul*=2;
        if(a[i][0]!=‘#‘&&a[i][m-1]!=‘#‘&&a[i][0]!=a[i][m-1])mul*=2;
        if(a[i][0]!=‘#‘)cnt[i]++;
        if(a[i][m-1]!=‘#‘)cnt[i]++;
        if(a[0][i]!=‘#‘)mask+=p[i-1];
        if(a[m-1][i]!=‘#‘)mask+=p[i-1];
    }
    dfs(0,0,0);
    f[0][mask][0]=mul;
    for(i=1;i<=n;i++){
        clr();
        int o=cnt[i];
        for(A=0;A<=mask;A++)f[e^1][A][o]=f[e][A][0];
        e^=1;
        for(j=0;j<n;j++)if(a[i][j+1]==‘.‘){
            clr();
            for(S=0;S<=mask;S++){
            		if(w[S]>>j&1)for(k=0;k<=o;k++)if(f[e][S][k])f[e^1][S-p[j]][k]+=f[e][S][k];
            		for(k=1;k<=o;k++)if(f[e][S][k])f[e^1][S][k-1]+=f[e][S][k];
            }
            e^=1;
        }
    }
    printf("%llu",f[e][0][0]);
}
/*
5
##ARRR#
H.#.##S
Y.....E
O.#.#.E
##....E
###.#.E
#XNT#A#
*/

M. Maximal paths

樹形DP求出每個點內部從葉子往上、從上往下的最大數字串即可。需要手寫高精度。

#include<cstdio>
#include<algorithm>
using namespace std;
typedef unsigned long long ll;
typedef __int128 lll;
const int N=10000010,M=50;
const ll MO=1000000000000000000ULL;
int n,i;char a[N];
struct P{
    ll v[3];
    P(){v[0]=v[1]=v[2]=0;}
    P operator+(int b){
        P c;
        for(int i=0;i<3;i++)c.v[i]=v[i];
        c.v[0]+=b;
        for(int i=0;i<2;i++)if(c.v[i]>=MO){
            c.v[i+1]+=c.v[i]/MO;
            c.v[i]-=MO;
        }
        return c;
    }
    P operator+(const P&b){
        P c;
        for(int i=0;i<3;i++)c.v[i]=v[i]+b.v[i];
        for(int i=0;i<2;i++)if(c.v[i]>=MO){
            c.v[i+1]+=c.v[i]/MO;
            c.v[i]-=MO;
        }
        return c;
    }
    P operator*(int b){
        P c;
        for(int i=0;i<3;i++)c.v[i]=v[i]*b;
        for(int i=0;i<2;i++)c.v[i+1]+=c.v[i]/MO,c.v[i]%=MO;
        c.v[2]%=MO;
        return c;
    }
    P operator*(const P&b){
        P c;
        static lll f[3];
        f[0]=f[1]=f[2]=0;
        for(int i=0;i<3;i++)if(v[i])for(int j=0;i+j<3;j++)if(b.v[j])f[i+j]+=(lll)v[i]*b.v[j];
        for(int i=0;i<2;i++)if(f[i]>=MO)f[i+1]+=f[i]/MO,f[i]%=MO;
        if(f[2]>=MO)f[2]%=MO;
        for(int i=0;i<3;i++)c.v[i]=f[i];
        return c;
    }
    void up(const P&b){//max=
        int i;
        for(i=2;~i;i--)if(b.v[i]!=v[i])break;
        if(i<0)return;
        if(b.v[i]<v[i])return;
        for(i=0;i<3;i++)v[i]=b.v[i];
    }
    void write(){
        int i=2;
        while(i&&!v[i])i--;
        printf("%llu",v[i]);
        for(int j=i-1;~j;j--)printf("%018llu",v[j]);
        puts("");
    }
}f[M],g[M],p[M],ans,ff,gg,mx[M];
int d[M];
unsigned int seed,base;
void dfs(int x,int k){
    f[k]=g[k]=mx[k]=P();
    d[k]=0;
    for(int i=0;i<2;i++){
        int y=x<<1|i;
        if(y>n)continue;
        dfs(y,k+1);
        ff=(f[k+1]*10)+((int)a[y]);
        gg=g[k+1]+(p[d[k+1]]*((int)a[y]));
        mx[k].up((ff*p[d[k]])+g[k]);
        mx[k].up((f[k]*p[d[k+1]+1])+gg);
        /*if(x==1){
            puts("debug");
            ff.write();
            f[k].write();
        }*/
        f[k].up(ff);
        g[k].up(gg);
        d[k]=max(d[k],d[k+1]+1);
        /*if(x==1){
            puts("debug");
            ff.write();
            f[k].write();
        }*/
    }
   /* printf("%d:\n",x);
    f[k].write();
    g[k].write();
    mx[k].write();*/
    ans=ans+mx[k];
}
int main(){
    freopen("input.txt","r",stdin);
    freopen("output.txt","w",stdout);
    scanf("%d%u",&n,&seed);
    base=(1U<<31)-1;
    p[0].v[0]=1;
    for(i=1;i<M;i++)p[i]=p[i-1]*10;
    for(i=1;i<=n;i++){
        a[i]=((seed&base)>>16)%9+1;
        seed=seed*1103515245+12345;
        //printf("%d\n",a[i]); i and i/2
    }
    dfs(1,0);
    ans.write();
}

XVII Open Cup named after E.V. Pankratiev. XXI Ural Championship