话说每天都在写水题啊，难题也不想写，算法也不想学，工程技术也不想学，就想着打游戏看小说，好颓废hh

狗和猫

超级水，看清题意模拟就好了

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#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;

int n, d, c, m;

signed main()
{
    ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);

    int t;
    cin >> t;
    for (int cas = 1; cas <= t; cas ++)
    {
        cin >> n >> d >> c >> m;
        string s;
        cin >> s;
        bool flag = true;
        for (int i = 0; i < s.size(); i ++)
        {
            if (s[i] == 'D')
                if(d)
                {
                    c += m;
                    d--;
                }
                else
                {
                    flag = false;
                    break;
                }
            else
                if (c)
                    c--;
                else
                {
                    for (int j = i; j < s.size(); j ++)
                        if (s[j] == 'D')
                        {
                            flag = false;
                        }
                    break;
                }
        }

        if (flag)
            cout << "Case #" << cas << ": " << "YES" << endl;
        else
            cout << "Case #" << cas << ": " << "NO" << endl;

    }
}

3326. 最大硬币数

由于每个格子里没有负数，所以直接累加对角线就行了,
// 主对角线的坐标, 坐标差值是相等的, 即 i - j 相同, 加n是为了防止负数出现
// 斜对角线的坐标 i + j 相同
这个处理方式在n皇后问题中出现过

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#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;

typedef long long LL;

const int N = 1005;
int a[N][N];
LL f[2 * N + 5]; // 对角线数组

signed main()
{
    ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
    
    int t;
    cin >> t;
    for (int cas = 1; cas <= t; cas ++)
    {
        int n;
        cin >> n;
        memset(f, 0, sizeof f);
        for (int i = 1; i <= n; i ++)
        {
            for (int j = 1; j <= n; j ++)
            {
                cin >> a[i][j];
                f[i - j + n] += a[i][j];
            }
        }
        LL maxn = 0;
        for (int i = 1; i <= 2 * n - 1; i ++ )
        {
            maxn = max(f[i], maxn);
        }
        
        cout << "Case #" << cas << ": " << maxn << endl;
    }
}