[JSOI2015]染色问题

标签：^c   简单   ace   代码   sum   cin   end   std   out   

一道比较简单的容斥题(?

可以容斥一下, 设\(i, j ,k\)分别代表\(i\)行有颜色, \(j\)行有颜色, 出现\(k\)种颜色的方案数
\[ ans = \sum_{i=0}^n\sum_{j=0}^m\sum_{k=0}^c(-1)^{(i+j+k+n+m+c)}C_n^iC_m^jC_c^k(k+1)^{(n*m)} \]
k+1是因为一个格子可以不填颜色, 反手一个预处理组合数, O(n^3)稳过

代码

#include<iostream>
#include<cstdio>
#include<cstring>
#define ll long long
using namespace std;
const int P = 1e9+7;
const int N = 505;
ll C[N][N], fpw[N*N];
int n, m, c;
ll ans;
int main() {
    cin >> n >> m >> c;
    fpw[0] = C[0][0] = 1;
    for (int i = 1;i <= 400; i++) {
        C[i][0] = 1;
        for (int j = 1;j <= i; j++) {
            C[i][j] = C[i-1][j] + C[i-1][j-1];
            if (C[i][j] >= P) C[i][j] -= P;
        }
    }
    int p = (n ^ m ^ c) & 1;
    for (int k = 0;k <= c; k++) {
        for (int i = 1;i <= n * m; i++) fpw[i] = fpw[i-1] * (k+1) % P;
        for (int i = 0;i <= n; i++) {
            for (int j = 0;j <= m; j++) {
                ll res = C[n][i] * C[m][j] % P * C[c][k] % P * fpw[i*j] % P;
                if (((i ^ j ^ k) & 1) ^ p) ans -= res;
                else ans += res;
            }
        }
    }
    cout << (ans % P + P) % P << endl;
    return 0;
}

[JSOI2015]染色问题

标签：^c   简单   ace   代码   sum   cin   end   std   out   

原文地址：https://www.cnblogs.com/Hs-black/p/12026679.html