HDU 5396 Expression(DP+组合数)（详解）

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
#include<vector>
#include<queue>
#include<cmath>
using namespace std;
#define INF 0x3fffffff
#define maxn 110

typedef long long LL;
const LL MOD = 1e9+7;

LL A[maxn], C[maxn][maxn];
char op[maxn];
LL dp[maxn][maxn];

int main()
{
    int n;
    A[0] = 1;
    for(int i=1; i<=100; i++)
        A[i] = (A[i-1] * i)%MOD;
    C[0][0] = 1;
    for(int i=1; i<=100; i++)
    {
        C[i][0] = 1;
        for(int j=1; j<=i; j++)
            C[i][j] = (C[i-1][j-1] + C[i-1][j])%MOD;
    }

    while(scanf("%d", &n) != EOF)
    {
        memset(dp, 0, sizeof(dp));
        for(int i=1; i<=n; i++)
            scanf("%I64d", &dp[i][i]);
        scanf("%s", op+1);

        for(int L=2; L <= n; L++)
        {
            for(int i=1; i+L-1 <= n; i++)
            {
                int j = i + L - 1;
                dp[i][j] = 0;
                for(int k=i; k<j; k++)
                {
                    LL t;
                    if(op[k] == ‘*‘)
                        t = (dp[i][k] * dp[k+1][j])%MOD;
                    if(op[k] == ‘+‘)
                        t = (dp[i][k]*A[j-k-1] + dp[k+1][j]*A[k-i])%MOD;
                    if(op[k] == ‘-‘)
                        t = (dp[i][k]*A[j-k-1] - dp[k+1][j]*A[k-i])%MOD;

                    dp[i][j] = (dp[i][j] + t * C[j-i-1][k-i])%MOD;
                }
            }
        }

        printf("%I64d\n", (dp[1][n]+MOD)%MOD );
    }
    return 0;
}