Number String

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1945 Accepted Submission(s): 938

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

题意：略。

思路：有技巧的dp，dp[i][j]表示处理到第i位数，以j结尾的方案数，如果s[i]为'I'，dp[i][j] = dp[i-1][j-1] + dp[i-1][j-2] + ... dp[i-1][1]。如果s[i]为'D'，dp[i][j] = dp[i-1][j+1] +dp[i-1][j+2] +...+dp[i-1][i]。

如果j在前面已经出现过了，那么可以理解为将前面>=j的数都加1，下面代码用前缀和表示。

# include 
    
     # include 
     
      # include 
      
       # define ll long long# define MOD 1000000007using namespace std;ll sum[1003][1003];int main(){    char s[1003];    while(~scanf("%s",s))    {        sum[0][1] = 1;        int len = strlen(s);        for(int i=1; i<=len; ++i)        {            for(int j=1; j<=i+1; ++j)            {                sum[i][j] = sum[i][j-1];                if(s[i-1] != 'D')                    sum[i][j] += sum[i-1][j-1];                if(s[i-1] != 'I')                    sum[i][j] += sum[i-1][i] - sum[i-1][j-1] + MOD;                sum[i][j] %= MOD;            }        }        printf("%lld\n",sum[len][len+1]);    }    return 0;}

转载于:https://www.cnblogs.com/junior19/p/6729891.html

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