HDU 4689 Derangement

DP.....

　　f[i][j]表示前 i 个数还有 j 个+号没有放数字，-号全部放满。

　　当 i 为+号时：1、当前这个数不放，即放在后面的位置中，f[i][j]+=f[i-1][j-1]。2、当前这个数放在前面的位置中，

f[i][j]+=f[i-1][j]*j。所以f[i][j]=f[i-1][j-1]+f[i-1][j]*j。

　　当 i 为-号时：1、当前这个数放在后面的位置中，由前面+号未填的数来填-号，f[i][j]+=f[i-1][j]*j。2、当前这个数

放在前面的位置中，前面 j 个空位对应的数放在当前的-号位，当前这个数放在前面未放的+号位，

f[i][j]+=f[i-1][j+1]*j*j。

Derangement

Time Limit: 7000/7000 MS (Java/Others)    Memory Limit: 65535/102400 K (Java/Others)
Total Submission(s): 838    Accepted Submission(s): 251

Problem Description

A derangement is a permutation such that none of the elements appear in their original position. For example, [5, 4, 1, 2, 3] is a derangement of [1, 2, 3, 4, 5]. Subtracting the original permutation from the derangement, we get the derangement difference [4, 2, -2, -2, -2], where none of its elements is zero. Taking the signs of these differences, we get the derangement sign [+, +, -, -, -]. Now given a derangement sign, how many derangements are there satisfying the given derangement sign?

 

Input

There are multiple test cases. Process to the End of File.
Each test case is a line of derangements sign whose length is between 1 and 20, inclusively.

 

Output

For each test case, output the number of derangements.

 

Sample Input

   
   
   
   

    
    
    
    +-
++---
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    1
13
   
   
   
   

 

Author

Zejun Wu (watashi)

 

Source

2013 Multi-University Training Contest 9

 

ll dp[N][N];
char s[N];
int main()
{
    while(scanf("%s",s)==1)
    {
        int n=strlen(s);
        memset(dp,0,sizeof dp);
        dp[0][0]=1;
        for(int k=0;k<=n;k++)
        {
            for(int i=0;i<=n;i++)
            {
                ll tmp = dp[k][i];
                if(tmp==0) continue;
              ///  printf("dp[%d][%d] = %d\n", k, i, dp[k][i]);
                if(s[k]=='+')
                {
                    dp[k+1][i+1] += tmp;
                    dp[k+1][i] += i * tmp;
                }
                if(s[k]=='-')
                {
                    dp[k+1][i] += 1ll * i * tmp;
                    if(i) dp[k+1][i-1] += 1ll * i * i * tmp;
                }
            }
           // putchar(10);
        }
        printf("%I64d\n", dp[n][0]);
    }
}