hdu 5396 区间dp+组合

http://acm.hdu.edu.cn/showproblem.php?pid=5396

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. 
    
   
   
   
   

/**
hdu 5396  区间dp+组合
题目大意：给定一个只含有+-*的表达式，让你加括号决定运算顺序，问所有运算顺序可得的总和是多少？
解题思路：看一眼就知道区间dp最后差一个地方没有调出来，悲剧了，看了题解才知道左右区间还要乘上组合数==
           （转） 用dp[l][r]表示第l个数到第r个数组成的各种顺序的表达式和是多少，t[l][r]表示第l个数到第r
           个数有多少种不同的组合。dp[l][r]的计算方法是枚举最后一个被计算的位置i，设n1=dp[l][i]，
           n2=dp[i+1][r]，t1=t[l][i]，t2=t[i+1][r]。那么对于加号，对于每个i要加上n1*t2+n2*t1，对于右边不同
           的组合，左边的数每次都要被加一次，同理左边不同的组合，右边的数每次也要被加一次。因此n1被加了t2次，
           n2被加了t1次。减法和加法一样。乘法是直接n1*n2。这还没完，注意就算是左边的顺序和右边的顺序的确定，
           假设左边有f1个符号，右边有f2个符号，也有C[f1+f2][f1]种排法，相当于在f1+f2个位置中选f1个，剩下的给f2，
           f1和f2中排列的相对顺序不改变，所以还要乘上C[f1+f2][f1]。同理对于每个i，t[l][r]要加上t1*t2*C[f1+f2][f1]。
*/
#include <string.h>
#include <algorithm>
#include <iostream>
#include <stdio.h>
using namespace std;
typedef long long LL;
const LL mod=1e9+7;
int n;
LL a[105],dp[105][105],num[105][105],c[105][105];
char s[105];

char getC()
{
    c[0][0]=1;
    for(int i=1;i<=100;i++)
    {
        c[i][0]=c[i][i]=1;
        for(int j=1;j<i;j++)
        {
            c[i][j]=(c[i-1][j]+c[i-1][j-1])%mod;
        }
    }
}
int main()
{
    getC();
    while(~scanf("%d",&n))
    {
        for(int i=1; i<=n; i++)
            scanf("%I64d",&a[i]);
        scanf("%s",s+1);
        memset(num,0,sizeof(num));
        memset(dp,0,sizeof(dp));
        for(int i=1; i<=n; i++)
        {
            num[i][i]=1;
            dp[i][i]=a[i];
        }
        for(int i=n-1; i>=1; i--)
        {
            for(int j=i+1; j<=n; j++)
            {
                LL cnt=0;
                for(int k=i; k<j; k++)
                {
                    if(s[k]=='*')
                    {
                        cnt=(dp[i][k]*dp[k+1][j])%mod;
                    }
                    else if(s[k]=='-')
                    {
                        cnt=((dp[i][k]*num[k+1][j])%mod-(dp[k+1][j]*num[i][k])%mod)%mod;
                    }
                    else
                    {
                        cnt=((dp[i][k]*num[k+1][j])%mod+(dp[k+1][j]*num[i][k])%mod)%mod;
                    }
                    dp[i][j]=(dp[i][j]+c[j-i-1][k-i]*cnt%mod)%mod;
                    num[i][j]=(num[i][j]+num[i][k]*num[k+1][j]%mod*c[j-i-1][k-i]%mod)%mod;
                }
                //printf("%d %d:%I64d\n",i,j,dp[i][j]);
            }
        }
        printf("%I64d\n",(dp[1][n]+mod)%mod);
    }
    return 0;
}