CF-Codeforces Round #420 (Div. 2) A ~ E

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ACM模版

A-Okabe and Future Gadget Laboratory

描述

题解

暴力搞搞，O(n4) 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>>>

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