Codeforces_821

A.直接判断每一个数。

#include<bits/stdc++.h>
using namespace std;

int n,a[][];

int main()
{
    ios::sync_with_stdio();
    cin >> n;
    for(int i = ;i <= n;i++)
    {
        for(int j = ;j <= n;j++)   cin >> a[i][j];
    }
    for(int i = ;i <= n;i++)
    {
        for(int j = ;j <= n;j++)
        {
            if(a[i][j] == )    continue;
            int flag = ;
            for(int ii = ;ii <= n;ii++)
            {
                for(int jj = ;jj <= n;jj++)
                {
                    if(a[ii][j]+a[i][jj] == a[i][j])    flag = ;
                }
            }
            if(!flag)
            {
                cout << "No" << endl;
                return ;
            }
        }
    }
    cout << "Yes" << endl;
    return ;
}

---

B.枚举直线上每一点，x，y坐标分开算，求和公式。

#include<bits/stdc++.h>
using namespace std;

long long m,b;

int main()
{
    ios::sync_with_stdio();
    cin >> m >> b;
    long long ans = ;
    for(int y = ;y <= b;y++)
    {
        long long x = (b-y)*m;
        ans = max(ans,(x+)*(y+)*y/+(y+)*(x+)*x/);
    }
    cout << ans << endl;
    return ;
}

---

C.模拟栈，每次需要整理的时候，把栈中所有出栈，因为底下的已经有序，不需要再去关注它们。

#include<bits/stdc++.h>
using namespace std;

int n;
stack<int> s;
string ss;

int main()
{
    ios::sync_with_stdio();
    cin >> n;
    int now = ,ans = ;
    for(int i = ;i <= *n;i++)
    {
        cin >> ss;
        if(ss == "add")
        {
            int x;
            cin >> x;
            s.push(x);
        }
        else
        {
            if(s.empty())
            {
                now++;
                continue;
            }
            if(now == s.top())
            {
                s.pop();
                now++;
            }
            else
            {
                ans++;
                now++;
                while(!s.empty())   s.pop();
            }
        }
    }
    cout << ans << endl;
    return ;
}

---

D.直接O(k^2)的最短路，转移的时候若两个点相邻，则dis不用+1，若某一坐标之差≤2，则可以转移dis+1，若终点不lit，设一个(n+1,m+1)的点，最后结果回自动+1。

#include<bits/stdc++.h>
#define INF 0x3f3f3f3f
using namespace std;

int n,m,k,x[],y[],vis[] = {},dis[] = {};

int main()
{
    ios::sync_with_stdio();
    cin >> n >> m >> k;
    int s,t = ;
    for(int i = ;i <= k;i++)
    {
        cin >> x[i] >> y[i];
        if(x[i] ==  && y[i] == )  s = i;
        if(x[i] == n && y[i] == m)  t = i;
    }
    if(!t)
    {
        t = ++k;
        x[k] = n+;
        y[k] = m+;
    }
    memset(dis,0x3f,sizeof(dis));
    dis[s] = ;
    for(int j = ;j <= k;j++)
    {
        int pos = -,minn = INF;
        for(int i = ;i <= k;i++)
        {
            if(!vis[i] && dis[i] < minn)
            {
                pos = i;
                minn = dis[i];
            }
        }
        if(pos == -)  break;
        vis[pos] = ;
        for(int i = ;i <= k;i++)
        {
            if(vis[i])  continue;
            if(abs(x[pos]-x[i])+abs(y[pos]-y[i]) == )  dis[i] = min(dis[i],dis[pos]);
            else if(abs(x[pos]-x[i]) <=  || abs(y[pos]-y[i]) <= ) dis[i] = min(dis[i],dis[pos]+);
        }
    }
    if(dis[t] == INF) cout << - << endl;
    else    cout << dis[t] << endl;
    return ;
}

---

E.对于每一段，可以dp，矩阵快速幂优化，注意每一段结束和每一段开始的时候，把高于c点的值清零。

#include<bits/stdc++.h>
#define MOD 1000000007
using namespace std;

int n;
long long k;
struct xx
{
    long long m[][];
};

xx mul(xx a,xx b,int len)
{
    xx tmp;
    for(int i = ;i <= len;i++)
    {
        for(int j = ;j <= len;j++)
        {
            tmp.m[i][j] = ;
            for(int k = ;k <= len;k++)  tmp.m[i][j] = (tmp.m[i][j]+a.m[i][k]*b.m[k][j])%MOD;
        }
    }
    return tmp;
}

xx qpower(xx a,long long k,int len)
{
    xx b;
    memset(b.m,,sizeof(b.m));
    for(int i = ;i <= len;i++)  b.m[i][i] = ;
    while(k)
    {
        if(k%) b = mul(a,b,len);
        a = mul(a,a,len);
        k /= ;
    }
    return b;
}
int main()
{
    ios::sync_with_stdio();
    cin >> n >> k;
    xx base,now;
    for(int i = ;i < ;i++)
    {
        for(int j = max(,i-);j < min(i+,);j++)  base.m[i][j] = ;
    }
    now.m[][] = ;
    for(int i = ;i <= n;i++)
    {
        long long a,b,c;
        cin >> a >> b >> c;
        for(int j = c+;j < ;j++) now.m[j][] = ;
        now = mul(qpower(base,min(b,k)-a,c),now,c);
        for(int j = c+;j < ;j++) now.m[j][] = ;
        if(b >= k)  break;
    }
    cout << now.m[][] << endl;
    return ;
}