Number String ZOJ - 3543
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.
这道dp题最巧妙的地方 感觉就在 那句 如果当前的位置放 x, 那么前面所有大于等于x的数可以看做同时+1，这样就是一个新排列了。 如果能想到这个等价性质，下面就比较好想了。 真是奇妙。
#include
using namespace std;
typedef long long LL;
#define rep(i,a,b) for(int i=a;i