hdu 3336 Count the string

Count the string

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
http://acm.hdu.edu.cn/showproblem.php?pid=3336

Problem Description

It is well known that AekdyCoin is good at string problems as well as number theory problems. When given a string s, we can write down all the non-empty prefixes of this string. For example:
s: "abab"
The prefixes are: "a", "ab", "aba", "abab"
For each prefix, we can count the times it matches in s. So we can see that prefix "a" matches twice, "ab" matches twice too, "aba" matches once, and "abab" matches once. Now you are asked to calculate the sum of the match times for all the prefixes. For "abab", it is 2 + 2 + 1 + 1 = 6.
The answer may be very large, so output the answer mod 10007.

 

Input

The first line is a single integer T, indicating the number of test cases.
For each case, the first line is an integer n (1 <= n <= 200000), which is the length of string s. A line follows giving the string s. The characters in the strings are all lower-case letters.

 

Output

For each case, output only one number: the sum of the match times for all the prefixes of s mod 10007.

 

Sample Input

1

4

abab

 

Sample Output

6

题目大意：求字符串前缀在字符串中出现的次数和

KMP+DP

首先求出字符串的next数组

状态转移方程：dp[i]+=dp[next[i]]+1,ans+=dp[i]

如何理解？

例：    a b a b a

next：0 0 1 2 3

dp：   1 1 2 2 3

对于+1，因为本身和本身匹配

 dp[3]=dp[1]+1 

因为next[3]=1,所以第1个和第3个是相同的

第1个也会出现在第3个中

只不过这里把它加到了第3个里

也就是说：

dp[i]=dp[j]+1

[1,j]=[i-j+1,i]

所以[1,j]中出现过的在[i-j+1,i]中重复出现

次数本应累加到[i,j]里，实际累加到了[i-j+1,i]里 

 

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
int t,n,len,dp[200001],ans;
char s[200001];
int f[200010];
void getnext()
{
	for(int i=1;i<n;i++)
	{
		int j=f[i];
		while(j&&s[i]!=s[j]) j=f[j];
		f[i+1]= s[i]==s[j] ? j+1:0;
	}
}
int main()
{
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d",&n);
		cin>>s;
		len=strlen(s);
		getnext();
		memset(dp,0,sizeof(dp));
		ans=0;
		for(int i=1;i<=n;i++)
		{
			dp[i]=dp[f[i]]+1;
			ans+=dp[i];
			dp[i]%=10007;
			ans%=10007;
		}
		printf("%d
",ans);
		/*for(int i=1;i<=n;i++)
		 printf("%d ",dp[i]); 
		printf("
");
		for(int i=1;i<=n;i++) printf("%d ",f[i]);*/
	}
}