[GCJ]Password Attacker

https://code.google.com/codejam/contest/4214486/dashboard#s=p0

排列组合。DP递推式，如下代码。dp[m][n]表示长度为n的字符串里有m个字符，那么可以先用m-1个字符拼一个长度为n-1的字符串，然后再C(n,1)里面挑一个放最后一个字符;这是最后一种字符是一个的情况，后面还有两个三个等等。所以代码如下：

要注意的是，可以先计算组合数combination[n][m]，用C(n,m)=C(n−1,m)+C(n−1,m−1)来算。

/*
f[i][n] = f[i-1][n-1]*C(n,1) + f[i-1][n02]*C(n,2) + ... + f[i-1][i-1] * C(n, n-(i-1));
*/

#include <iostream>
#include <vector>
using namespace std;

int base = 1000000007;
typedef long long llong;

llong combination[101][101];

void buildCombination() {
    for (int i = 0; i <= 100; i++) {
        for (int j = 0; j <= i; j++) {
            if (j == 0) {
                combination[i][j] = 1;
            } else {
                combination[i][j] = (combination[i-1][j] + combination[i-1][j-1]) % base;
            }
        }
    }
}

llong solve(int m, int n) {
    vector<vector<llong> > dp;
    dp.resize(m+1);
    for (int i = 0; i < m+1; i++) {
        dp[i].resize(n+1);
    }
    // i chars, len of j
    for (int i = 1; i <= m; i++) {
        for (int j = i; j <= n; j++) {
            if (i == 1) {
                dp[i][j] = 1;
                continue;
            }
            dp[i][j] = 0;
            for (int k = 1; j-k >= i-1; k++) {
                dp[i][j] = (dp[i][j] + dp[i-1][j-k] * combination[j][k]) % base;

            }
        }
    }
    return dp[m][n];
}

int main() {    
    int T;
    buildCombination();
    cin >> T;
    for (int i = 1; i <= T; i++) {
        int m, n;
        cin >> m >> n;
        llong r = solve(m, n);
        cout << "Case #" << i << ": " << r << endl;       
    }
    return 0;
}