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：
- '+' 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)!
- '-' 与'+'类似
- '*' 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; }