hdu 4055 Number String (基础dp）

Number String

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1935    Accepted Submission(s): 931

Problem Description

The signature of a permutation is a string that is computed as follows: for each pair of consecutive elements of the permutation, write down the letter 'I' (increasing) if the second element is greater than the first one, otherwise write down the letter 'D' (decreasing). For example, the signature of the permutation {3,1,2,7,4,6,5} is "DIIDID".

Your task is as follows: You are given a string describing the signature of many possible permutations, find out how many permutations satisfy this signature.

Note: For any positive integer n, a permutation of n elements is a sequence of length n that contains each of the integers 1 through n exactly once.

 

Input

Each test case consists of a string of 1 to 1000 characters long, containing only the letters 'I', 'D' or '?', representing a permutation signature.

Each test case occupies exactly one single line, without leading or trailing spaces.

Proceed to the end of file. The '?' in these strings can be either 'I' or 'D'.

 

Output

For each test case, print the number of permutations satisfying the signature on a single line. In case the result is too large, print the remainder modulo 1000000007.

 

Sample Input

II ID DI DD ?D ??

 

Sample Output

1 2 2 1 3 6

Hint

Permutation {1,2,3} has signature "II". Permutations {1,3,2} and {2,3,1} have signature "ID". Permutations {3,1,2} and {2,1,3} have signature "DI". Permutation {3,2,1} has signature "DD". "?D" can be either "ID" or "DD". "??" gives all possible permutations of length 3.

 

Author

HONG, Qize

---

基础dp

 

字符串长度为n，则有（n+1）个数字组成排列。

dp[i][j]代表长度为i末位为j的符合题意的全排列总数，所以这样的排列内没有大于i的数。

显然dp[1][1]=1;

对于i>1,若对应的字符为‘I’,那么$dp[i][j]=sum_{k=1}^{j-1}dp[i-1][k]$，这个显而易见。

若对应的字符为'D'，那么$dp[i][j]=sum_{k=j}^{i-1}dp[i-1][k]$，这个其实相当于在i位置放入j以后，把i全排列中i-1及之前的位置大于j的数字+1，构成新的排列。

那么对应的字符为“?”，则是把上两种情况综合，$dp[i][j]=sum_{k=1}^{i-1}dp[i-1][k]$，显而易见。

#include<cstdio>
#include<iostream>
#include<cstring>
#define clr(x) memset(x,0,sizeof(x))
#define LL long long
#define mod 1000000007
using namespace std;
LL dp[1010][1010],ans;
char s[1010];
int main()
{
    while(scanf("%s",s)!=EOF)
    {
        dp[1][1]=1;
        for(int i=2;i<=strlen(s)+1;i++)
        {
            if(s[i-2]=='I')
            {
                dp[i][1]=0;
                for(int j=2;j<=i;j++)
                    dp[i][j]=(dp[i][j-1]+dp[i-1][j-1])%mod;
            }
            if(s[i-2]=='D')
            {
                dp[i][i]=0;
                for(int j=i-1;j>=1;j--)
                    dp[i][j]=(dp[i][j+1]+dp[i-1][j])%mod;
            }
            if(s[i-2]=='?')
            {
                dp[i][1]=0;
                for(int j=1;j<=i-1;j++)
                    dp[i][1]=(dp[i][1]+dp[i-1][j])%mod;
                for(int j=2;j<=i;j++)
                    dp[i][j]=dp[i][j-1];
            }
        }
        ans=0;
        for(int i=1;i<=strlen(s)+1;i++)
            ans=(ans+dp[strlen(s)+1][i])%mod;
        printf("%lld
",ans);
    }
    return 0;
}