http://acm.hdu.edu.cn/showproblem.php?pid=3336
Count the string
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 17068 Accepted Submission(s): 7721
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
Sample Output
Author
foreverlin@HNU
Source
Recommend
lcy
题意：求字符串所有前缀出现的次数（包括本身）
思路：求所有next的值的个数加上前缀本身
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <algorithm>
#include <iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include <stdio.h>
#include <string.h>
using namespace std;
char a[200009];
int num;
void getnext(char* a, int len , int *next)
{
next[0] = -1 ;
int k = -1 , j = 0 ;
while(j < len)
{
if(k == -1 || a[j] == a[k])
{
k++;
j++;
// if(a[j] != a[k])
next[j] = k ;
// else
// {
// next[j] = next[k];
// }
}
else
{
k = next[k];
}
}
}
int main()
{int n ;
scanf("%d" , &n);
while(n--)
{
int next[200009];
int l ;
scanf("%d" , &l);
scanf("%s" , a);
getnext(a , l , next);
int j = 0 ;
for(int i = 0 ; i <= l ; i++)
{
// cout << next[i] << " " ;
j = i ;
while(next[j] > 0)
{
num = (num + 1) % 10007 ;
j = next[j];
}
}
// cout << endl ;
printf("%d\n" , (num + l)%10007);
num = 0 ;
}
return 0 ;
}
kmp（前缀出现次数next应用）
原文：https://www.cnblogs.com/nonames/p/11296242.html
