HDU 4055 Number String (dp的思想)
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Number String
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1052 Accepted Submission(s): 456
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
IIIDDIDD?D??
Sample Output
122136HintPermutation {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
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#include <iostream>#include <cstring>#include <cstdio>#include <string>using namespace std;const int maxn=1010;const int mod=1000000007;string st;long long dp[maxn][maxn],sum[maxn];void computing(){int len=st.length();dp[1][1]=1;sum[1]=1;sum[0]=0;for(int i=2;i<=1+len;i++){for(int j=1;j<=i;j++){if(st[i-2]=='D') dp[i][j]=sum[j-1];else if(st[i-2]=='I') dp[i][j]=((sum[i-1]-sum[j-1])%mod+mod)%mod;else dp[i][j]=sum[i-1];}sum[0]=0;for(int j=1;j<=i;j++){sum[j]=(sum[j-1]+dp[i][j])%mod;}}cout<<sum[len+1]<<endl;}int main(){while(cin>>st){computing();}return 0;}
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