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题目链接：uva 12075 - Counting Triangles

题目大意：一个n?m的矩阵，求说有选任意三点，可以组成多少个三角形。

解题思路：任意选三点C(3(n+1)?(m+1))但是有些组合是不可行得，即为三点共线，除了水平和竖直上的组合，就是斜线上的了，dp[i][j]即为ij情况下的斜线三点共线。

#include <cstdio>
#include <cstring>

typedef long long ll;

const int N = 1005;
ll dp[N][N];

ll gcd (ll a, ll b) {
    return b == 0 ? a : gcd(b, a%b);
}

void init () {
    for (int i = 2; i < N; i++)
        for (int j = 2; j < N; j++)
            dp[i][j] = dp[i-1][j] + dp[i][j-1] - dp[i-1][j-1] + gcd(i, j) - 1;

    for (int i = 2; i < N; i++)
        for (int j = 2; j < N; j++)
            dp[i][j] += dp[i-1][j] + dp[i][j-1] - dp[i-1][j-1];
}

ll C(ll n, ll m) {
    if (n < m)
        return 0;

    ll ans = 1;
    for (ll i = 0; i < m; i++)
        ans = ans * (n-i) / (i+1);
    return ans;
}

ll solve (ll n, ll m) {
    return C((n)*(m), 3) - C(n, 3) * m - C(m, 3) * n;
}

int main () {
    int cas = 1;
    ll n, m;
    init();
    while (scanf("%lld%lld", &n, &m) == 2 && n && m) {
        printf("Case %d: %lld\n", cas++, solve(n+1, m+1)-dp[n][m]*2);
    }
    return 0;
}

uva 12075 - Counting Triangles(容斥原理),布布扣,bubuko.com

uva 12075 - Counting Triangles(容斥原理)

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原文地址：http://blog.csdn.net/keshuai19940722/article/details/26811165