【区间DP】 HDU 5396 Expression

通道

题意：给出n个数，n-1个操作方法，求(n-1)!的操作顺序得到的和是多少

思路：

 

代码：

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

typedef long long ll;

template <class T>
inline bool rd(T &ret) {
    char c; int sgn;
    if(c = getchar() , c == EOF) return false;
    while(c != '-' && (c < '0' || c > '9')) c = getchar();
    sgn = (c == '-') ? -1 : 1;
    ret = (c == '-') ? 0 : (c - '0');
    while(c = getchar(), c >= '0' && c <= '9') ret = ret * 10 + (c - '0');
    ret *= sgn;
    return true;
}

const int N = 101;
const int MOD = 1e9 + 7;

int n;
int a[N];
ll dp[N][N], f[N], C[N][N];
char op[N];

int main() {
    f[0] = 1;
    C[0][0] = 1;
    for (int i = 1; i < N; ++i) {
        f[i] = f[i - 1] * i % MOD;
        C[i][0] = C[i][i] = 1;
        for (int j = 1; j < i; ++j)
            C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % MOD;
    }
    while(1 == scanf("%d", &n)) {
        for (int i = 1; i <= n; ++i) rd(a[i]);
        scanf("%s", op + 1);
        memset(dp, 0, sizeof dp);
        for (int i = 1; i <= n; ++i) dp[i][i] = a[i];
        for(int l = 2; l <= n; ++l) 
            for(int i = 1; i <= n - l + 1; ++i) {
                int j = i + l - 1;
                ll now = 0;
                for(int k = i; k < j; ++k) {
                    if (op[k] == '*') now = dp[i][k] * dp[k + 1][j] % MOD;
                    else if (op[k] == '+') 
                        now = (f[j - k - 1] * dp[i][k] % MOD + f[k - i] * dp[k + 1][j] % MOD) % MOD;
                    else 
                        now = (f[j - k - 1] * dp[i][k] % MOD - f[k - i] * dp[k + 1][j] % MOD + MOD) % MOD;
                        dp[i][j] = (dp[i][j] + now * C[l - 2][k - i]) % MOD;
                }
                
            }
        printf("%d\n", dp[1][n]);
    }
    return 0;
}