hdu 5396 Expression 2015多校联合训练赛#9 动态规划 dp

Expression

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 372    Accepted Submission(s): 209

Problem Description

Teacher Mai has  n  numbers  a1,a2,⋯,an and  n−1  operators("+", "-" 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

There are multiple test cases.

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 "+","-" and "*", which represents the operator sequence.

 

Output

For each test case print the answer modulo  109+7 .

 

Sample Input

   
   
   
   

    
    
    
    3
3 2 1
-+
5
1 4 6 8 3
+*-*
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    2
999999689

    
    
    
    
 
     
 
      Hint 
     
 Two numbers are considered different when they are in different positions. 
    
    
    
    
     
   
   
   
   

 

Author

xudyh

 

Source

2015 Multi-University Training Contest 9

 

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
using namespace std;
#define ll long long
#define maxn 102
int mod = 1000000007;
ll dp[maxn][maxn];
int num[maxn];
char word[maxn];
ll C[maxn][maxn];
ll A[maxn];
int main(){
    int n;
    memset(C,0,sizeof(C));
    A[0] = 1;
    for(int i = 1;i < maxn; i++)
        A[i] = A[i-1]*i%mod;
    for(int i = 0;i < maxn; 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;
    }
    ll res = 0;
    while(scanf("%d",&n)!=EOF){
        for(int i = 0;i < n; i++)
            scanf("%d",&num[i]);
        scanf("%s",word);
        memset(dp,0,sizeof(dp));
        for(int i = 0;i < n; i++)
            dp[i][i] = num[i];
        for(int i = 1;i < n; i++){
            for(int j = 0;j + i < n; j++){
                for(int k = j;k < j+i; k++){
                    if(word[k] == '*')
                        res = dp[j][k]*dp[k+1][j+i]%mod;
                    else if(word[k] == '+')
                        res = (dp[j][k]*A[j+i-k-1]%mod+dp[k+1][j+i]*A[k-j]%mod);
                    else
                        res = (dp[j][k]*A[j+i-k-1]%mod-dp[k+1][j+i]*A[k-j]%mod);

                    dp[j][j+i] = (dp[j][j+i]+res*C[i-1][k-j])%mod;
                }
            }
        }
        dp[0][n-1] = (dp[0][n-1]%mod+mod)%mod;
        cout<<dp[0][n-1]<<endl;
    }
    return 0;
}