B - Number String

最新推荐文章于 2019-07-13 14:51:00 发布

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本文介绍了一种算法，用于计算满足特定排列签名的所有可能排列的数量。排列签名由一系列'I'（递增）和'D'（递减）组成，代表排列中元素的相对顺序变化。文章提供了一个示例程序，演示了如何通过动态规划的方法高效地解决这一问题。

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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

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.

#include<stdio.h>
#include<algorithm>
#include<iostream>
#include <cstring>
using namespace std;
#define MAXN 1000
#define mod 1000000007
#define INF 0x7fffffff
typedef long long LL;
LL dp[MAXN+10][MAXN+10],sum[MAXN+10][MAXN+10];
char a[MAXN+10];
int main()
{
    while(scanf("%s",a)!=EOF)
    {
        int n=strlen(a)+1;
		dp[1][1]=sum[1][1]=1;
		for(int i=2;i<=n;i++)
		{
			for(int j=1;j<=i;j++)
			{
				if(a[i-2]=='D') dp[i][j]=sum[i-1][i-1]-sum[i-1][j-1];
				if(a[i-2]=='I') dp[i][j]=sum[i-1][j-1];
				if(a[i-2]=='?') dp[i][j]=sum[i-1][i-1];
				sum[i][j]=(sum[i][j-1]+dp[i][j])%mod;
			}
		}
		printf("%lld\n",(sum[n][n]+mod)%mod);
    }
    return 0;
}