HDU 4055 Number String （dp的思想）

最新推荐文章于 2019-11-05 00:42:39 发布

转载最新推荐文章于 2019-11-05 00:42:39 发布 · 116 阅读

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原文链接：http://www.cnblogs.com/toyking/p/3797385.html

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#java

本文介绍了一个关于排列签名的问题，即给定一个描述排列可能签名的字符串，找出满足该签名的所有排列数量。文章提供了完整的代码实现，包括输入字符串的处理、动态规划算法的应用以及输出结果的计算。

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

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.

Author

HONG, Qize

Source

2011 Asia Dalian Regional Contest

Recommend

lcy

#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;
}

转载于:https://www.cnblogs.com/toyking/p/3797385.html