字符串3

Count the string

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 5   Accepted Submission(s) : 3

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

14abab

 

 

Sample Output

6

 

 

Author

foreverlin@HNU

 

 

Source

HDOJ Monthly Contest – 2010.03.06

 

 

题意：给出一个字符串，问包含所有前缀的个数。

思路：kmp算法中的next数组的运用加dp的思想吧。

#include<iostream>#include<stdio.h>#include<string.h>using namespace std;char str[200005];int next[200005];int dp[200005];//dp[i]表示以str[i]结尾的所包含子串的个数。//next[i]表示以str[i]结尾的后缀与前缀的匹配的最大个数。void getNext(char *b,int m){    int tmp=0;    memset(next,0,sizeof(next));    for(int i=1;i<m;i++)    {        int tmp=next[i-1];        while(tmp&&b[tmp]!=b[i])        tmp=next[tmp-1];        if(b[tmp]==b[i])        next[i]=tmp+1;        else        next[i]=0;    }}int main(){    int t,n;    scanf("%d",&t);    while(t--)    {        int sum=0;        memset(dp,0,sizeof(dp));        scanf("%d",&n);        getchar();        for(int i=0;i<n;i++)        scanf("%c",&str[i]);        getNext(str,n);        for(int i=1;i<=n;i++)        {            dp[i]=dp[next[i-1]]+1;//因为以上次匹配位置i到本次之间的字符j为结尾的字符串不能匹配                                  //否则next[j]!=i;应该会是i到j之间的数。            sum=(sum+dp[i])%10007;        }        printf("%d\n",sum);    }    return 0;}