[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;

}