HDU 4055 Number String

Number String

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

 

Source

2011 Asia Dalian Regional Contest

 

题意：给出一组字符串，对于每一组字符串，若其长度为n，则在求n+1个数的全排列时，算出满足字符串要求的排列总数 mod 1000000007的值。其中I表示增长，D表示下降。即要满足数列中第i个位置与第i+1个位置的数字必须满足字符串中第i个字母的要求。

 

题解：动态规划。

这道题很明显是一道DP了吧。

f[i][j]表示长度为i的全排列，符合字符串的要求，最后一位数字是j的方案数。

于是对于increase(I)要求，f[i+1][j]+=f[i][1..j-1]；对于down(D)要求，f[i+1][j]+=f[i][j..i]；对于‘？’操作，我们之需要把increase和down并在一起即可。

有人可能理解不了down部分的转移方程。对于increase部分很显然是在前一位是1到j-1的情况下增加数j。但是按照increase的理解，那down部分岂不是要理解成在j到i的后面补上j？这样j元素就会重复一遍。

其实不然，我们用了转换的思想，我们可以把j到i看成从j+1到i+1。这样的话相当于把原序列的最后一个数加上1，此时既不会影响前面处理好的部分，还可以帮助后面转移。

很显然，我们需要维护一个前缀和sum[i][j]，将O(n³)的复杂度降到O(n²)的复杂度。

总方案数是f[i+1][1..i+1]，即sum[i+1][i+1]的值。

#include<bits/stdc++.h>
#define mod 1000000007
using namespace std;
const int N=1005;
long long f[N][N],sum[N][N];
char s[N];
int main()
{
    while (scanf("%s",s+1)!=EOF)
    {
        int n=strlen(s+1);
        f[1][1]=sum[1][1]=1;
        for (int i=1;i<=n;++i)
        for (int j=1;j<=i+1;++j)
        {
            if (s[i]=='I') f[i+1][j]=sum[i][j-1];
            else if (s[i]=='D') f[i+1][j]=(sum[i][i]-sum[i][j-1]+mod)%mod;
            else f[i+1][j]=sum[i][i];
            sum[i+1][j]=(sum[i+1][j-1]+f[i+1][j])%mod;
        }
        printf("%lld
",sum[n+1][n+1]);
    }
    return 0;
}