ACM模版

A-Okabe and Future Gadget Laboratory

描述

题解

暴力搞搞， O(n4) O ( n 4 ) 水过。

代码

#include <iostream>
#include <cstdio>

using namespace std;

const int MAXN = 55;

int n;
int A[MAXN][MAXN];
int vis[MAXN][MAXN];

int main(int argc, const char * argv[])
{
    cin >> n;
    for (int i = 0; i < n; i++)
    {
        for (int j = 0; j < n; j++)
        {
            scanf("%d", A[i] + j);
        }
    }

    for (int x = 0; x < n; x++)
    {
        for (int s = 0; s < n; s++)
        {
            int t1 = A[x][s];
            for (int t = 0; t < n; t++)
            {
                for (int y = 0; y < n; y++)
                {
                    int t2 = A[t][y];
                    if (t1 + t2 == A[x][y])
                    {
                        vis[x][y] = 1;
                    }
                }
            }
        }
    }

    int flag = 1;
    for (int i = 0; i < n; i++)
    {
        for (int j = 0; j < n; j++)
        {
            if (!vis[i][j] && A[i][j] != 1)
            {
                flag = 0;
                break;
            }
        }
        if (!flag)
        {
            break;
        }
    }

    if (flag)
    {
        cout << "Yes\n";
    }
    else
    {
        cout << "No\n";
    }

    return 0;
}

B-Okabe and Banana Trees

描述

题解

扫描一遍，暴力水过，这里建议扫描纵轴，因为横轴真的太长了，纵轴就短很多了，利用等差公式快速求解即可。

代码

#include <iostream>

using namespace std;

typedef long long ll;

ll m, b;

int main(int argc, const char * argv[])
{
    cin >> m >> b;

    ll ans = 0;
    for (int i = 0; i <= b; i++)
    {
        ll x = (b - i) * m;
        ll s = i * (i + 1) >> 1;
        ll e = s + x * (i + 1);
        ans = max(ans, (s + e) * (x + 1) >> 1);
    }

    cout << ans << '\n';

    return 0;
}

C-Okabe and Boxes

模拟，STL。详解>>>

D-Okabe and City

最短路，难在建图。详解>>>

E-Okabe and El Psy Kongroo

矩阵快速幂优化的动态规划问题。详解>>>