[hdu5396 Expression]区间DP

题意：给一个表达式，求所有的计算顺序产生的结果总和

思路：比较明显的区间dp，令dp[l][r]为闭区间[l,r]的所有可能的结果和，考虑最后一个符号的位置k，k必须在l,r之间，则l≤k<r，dp[l][r]=Σ{dp[l][k]?dp[k+1][r]}*(r-l-1)!/[(k-l)!(r-k-1)!]，其中(r-l-1)!/[(k-l)!(r-k-1)!]表示从左区间和右区间选择符号的不同方法总数（把左右区间看成整体，那么符号的选择在整体间也有顺序，内部的顺序不用管，那是子问题需要考虑的)，相当于(k-l)个0和(r-k-1)个1放一起的不同排列方法总数。

对花括号里面的'?'分为三种情况：

1. '+'  假设左区间有x种可能的方法，右区间有y种可能的方法，由于分配律的存在，左边的所有结果和会重复计算y次，右边的所有结果和会重复计算x次，而左边共(k-l)个符号，右边共(r-k-1)个符号，所以合并后的答案dp[l][r]=dp[l][k]*(r-k-1)!+dp[k+1][r]*(k-l)!
2. '-'   与'+'类似
3. '*'   由分配律，合并后的答案dp[l][r]=dp[l][k]*dp[k+1][r]

#pragma comment(linker, "/STACK:10240000")
#include <map>
#include <set>
#include <cmath>
#include <ctime>
#include <deque>
#include <queue>
#include <stack>
#include <vector>
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

#define X                   first
#define Y                   second
#define pb                  push_back
#define mp                  make_pair
#define all(a)              (a).begin(), (a).end()
#define fillchar(a, x)      memset(a, x, sizeof(a))
#define copy(a, b)          memcpy(a, b, sizeof(a))

typedef long long ll;
typedef pair<int, int> pii;
typedef unsigned long long ull;

//#ifndef ONLINE_JUDGE
void RI(vector<int>&a,int n){a.resize(n);for(int i=0;i<n;i++)scanf("%d",&a[i]);}
void RI(){}void RI(int&X){scanf("%d",&X);}template<typename...R>
void RI(int&f,R&...r){RI(f);RI(r...);}void RI(int*p,int*q){int d=p<q?1:-1;
while(p!=q){scanf("%d",p);p+=d;}}void print(){cout<<endl;}template<typename T>
void print(const T t){cout<<t<<endl;}template<typename F,typename...R>
void print(const F f,const R...r){cout<<f<<", ";print(r...);}template<typename T>
void print(T*p, T*q){int d=p<q?1:-1;while(p!=q){cout<<*p<<", ";p+=d;}cout<<endl;}
//#endif
template<typename T>bool umax(T&a, const T&b){return b<=a?false:(a=b,true);}
template<typename T>bool umin(T&a, const T&b){return b>=a?false:(a=b,true);}

const double PI = acos(-1.0);
const int INF = 1e9 + 7;
const double EPS = 1e-12;

/* -------------------------------------------------------------------------------- */

template<int mod>
struct ModInt {
    const static int MD = mod;
    int x;
    ModInt(ll x = 0): x(x % MD) {}
    int get() { return x; }

    ModInt operator + (const ModInt &that) const { int x0 = x + that.x; return ModInt(x0 < MD? x0 : x0 - MD); }
    ModInt operator - (const ModInt &that) const { int x0 = x - that.x; return ModInt(x0 < MD? x0 + MD : x0); }
    ModInt operator * (const ModInt &that) const { return ModInt((long long)x * that.x % MD); }
    ModInt operator / (const ModInt &that) const { return *this * that.inverse(); }

    ModInt operator += (const ModInt &that) { x += that.x; if (x >= MD) x -= MD; }
    ModInt operator -= (const ModInt &that) { x -= that.x; if (x < 0) x += MD; }
    ModInt operator *= (const ModInt &that) { x = (long long)x * that.x % MD; }
    ModInt operator /= (const ModInt &that) { *this = *this / that; }

    ModInt inverse() const {
        int a = x, b = MD, u = 1, v = 0;
        while(b) {
            int t = a / b;
            a -= t * b; std::swap(a, b);
            u -= t * v; std::swap(u, v);
        }
        if(u < 0) u += MD;
        return u;
    }

};
typedef ModInt<1000000007> mint;

const int maxn = 1e2 + 7;
const int md = 1e9 + 7;
mint dp[maxn][maxn], fac[maxn], facinv[maxn];
int a[maxn];
char s[maxn];

mint get(mint a, char ch, mint b, mint ca, mint cb) {
    if (ch == '+') return a * cb + b * ca;
    if (ch == '-') return a * cb - b * ca;
    if (ch == '*') return a * b;
}

void init() {
    fac[0] = facinv[0] = 1;
    for (int i = 1; i < maxn; i ++) fac[i] = fac[i - 1] * i;
    for (int i = 1; i < maxn; i ++) facinv[i] = facinv[i - 1] / i;
}

int main() {
#ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
#endif // ONLINE_JUDGE
    int n;
    init();
    while (cin >> n) {
        for (int i = 1; i <= n; i ++) {
            scanf("%d", a + i);
        }
        scanf("%s", s);
        fillchar(dp, 0);
        for (int i = 1; i <= n; i ++) dp[i][i] = a[i];
        for (int L = 2; L <= n; L ++) {
            for (int i = 1; i + L - 1 <= n; i ++) {
                int j = i + L - 1;
                for (int k = i; k < j; k ++) {
                    dp[i][j] += get(dp[i][k], s[k - 1], dp[k + 1][j], fac[k - i], fac[j - k - 1]) * fac[j - i - 1] * facinv[k - i] * facinv[j - k - 1];
                }
            }
        }
        printf("%d
", dp[1][n].get());
    }
    return 0;
}