Petrozavodsk Summer Training Camp 2015 Day 2: Xudyh (TooSimple) Contest, Saturday, August 22, 2015 A题

Problem A.

Expression Input file: standard input

Output file: standard output

Time limit: 1 second

Memory limit: 64 mebibytes Teacher Mai has n numbers a1, a2, . . . , an and n − 1 operators (each operator is one of ‘+’, ‘-’ or ‘*’) op1, op2, . . . , opn−1, which are arranged in the form a1 op1 a2 op2 a3 . . . an. He wants to erase numbers one by one. In i-th round, there are n + 1 − i numbers remained. He can erase two adjacent numbers and the operator between them, and then put a new number (derived from this one operation) in this position. After n − 1 rounds, there is the only one number remained. The result of this sequence of operations is the last number remained. He wants to know the sum of results of all different sequences of operations. Two sequences of operations are considered different if and only if in one round he chooses different numbers. For example, a possible sequence of operations for 1 + 4 ∗ 6 − 8 ∗ 3 is 1 + 4 ∗ 6 − 8 ∗ 3 → 1 + 4 ∗ (−2) ∗ 3 → 1 + (−8) ∗ 3 → (−7) ∗ 3 → −21. Input First line of the input contains one integer T (1 ≤ T ≤ 20) — number of the test cases. Then T test cases follow. For each test case, the first line contains one number n (2 ≤ n ≤ 100). The second line contains n integers a1, a2, . . . , an (0 ≤ ai ≤ 109 ). The third line contains a string with length n−1 consisting of ‘+’, ‘-’ and ‘*’, which represents the operator sequence. Output For each test case print the answer modulo 109 + 7. Example standard input standard output 2 3 3 2 1 -+ 5 1 4 6 8 3 +*-* 2 999999689

题目分析：考虑区间dp dp[i][j]表示i到j的所有可能的组合的和， 考虑转移，<i,j>可以由sigam<i,k>&<k,j>转移;

对于+或-号，dp[i][j]+=(dp[i][k]*fac[j-k]+dp[k+1][j]*fac[j-k-1])*C[k-i][j-i];

对于*号 dp[i][j]+=(dp[i][k]*dp[k+1][j])*C[k-i][j-i];

#include<bits/stdc++.h>
#define maxn 310
#define M 100
#define LL long long
#define Mod 1000000007 
using namespace std;
LL fac[maxn],C[maxn][maxn],dp[maxn][maxn];
void pre(){
    fac[0]=1;
    for(int i=1;i<=M;i++){
        fac[i]=(fac[i-1]*i)%Mod;
    }
    C[0][0]=1;
    for(int i=1;i<=M;i++){
        C[i][i]=C[i][0]=1;
        for(int j=1;j<i;j++){
            C[i][j]=(C[i-1][j]+C[i-1][j-1])%Mod;
        }
    }
}
char op[maxn];
LL seg[maxn];
int n,m;
int main(){
    pre();
    int T;
    cin>>T;
    while(T--){
    cin>>n;
    memset(dp,0,sizeof(dp));
    for(int i=1;i<=n;i++){
        cin>>dp[i][i];
    }
    cin>>op;
    m=strlen(op);
    for(int k=1;k<=n;k++){
        for(int i=1;i+k<=n;i++){
            for(int j=i;j<i+k;j++){
                if(op[j-1]=='*'){
                    dp[i][i+k]=(dp[i][i+k]+(((Mod+dp[i][j]*dp[j+1][i+k])%Mod)*C[k-1][j-i])%Mod)%Mod;
                }    
                else if(op[j-1]=='+'){
                    dp[i][i+k]=(dp[i][i+k]+(2*Mod+(dp[i][j]*fac[i+k-j-1])%Mod+(dp[j+1][i+k]*fac[j-i])%Mod)*C[k-1][j-i])%Mod;
                }
                else if(op[j-1]=='-'){
                    dp[i][i+k]=(dp[i][i+k]+((2*Mod+(dp[i][j]*fac[i+k-j-1])%Mod-(dp[j+1][i+k]*fac[j-i]))%Mod)*C[k-1][j-i])%Mod;
                }
            }
        }
    }
    cout<<(dp[1][n]+Mod)%Mod<<endl;
    }
    return 0;
}