Count the string

Total Submission(s) : 22   Accepted Submission(s) : 11

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

 

Author

foreverlin@HNU

 

Source

HDOJ Monthly Contest – 2010.03.06

 

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这个问题我刚开始想用循环做，就用kmp算法，对于字符串慢慢求出它的子字符串的出现次数，加起来得出最后结果，但是遗憾的是运算太过于复杂，导致TL

然后痛定思痛，在百度上找到了这个算法

感谢亲爱的博主：

这题很纠结，一开始不知道怎么做，如果直接统计子串在主串中出现的次数，orz···肯定TLE，后来发现这题可以直接从next数组入手，因为next数组表示的是子串中最长公共前后缀串的长度，如果用dt[i]表示该字符串前i个字符中出现任意以第i个字符结尾的前缀的次数，它的递推式是 dt[i]=d[next[i]]+1,即以第i个字符结尾的前缀数等于以第next[i]个字符为结尾的前缀数加上它自己本身，这里要好好理解一下，不太好解释。
举个例子：
              i  1 2 3 4 5 6
        字符串  a b a b a b
        dt[i]  1 1 2 2 3 3
        aba中出现的前缀为a,aba,所以dt[3]是2，ababa中出现的前缀为a,aba,ababa,所以dt[5]是3，当i=5时,next[5]=3，所以dt[i]=dt[next[i]]+1
理解了上面的部分就很简单了，后面直接套next数组的模板

#include <stdio.h>#include <stdlib.h>#include<string.h>char a[100000005];int next[100000005];int dt[100000005];void getnext(char *p){    int j,k;    int i=0;    int len=strlen(p);    next[0]=-1;    j=0;    k=-1;    while(j<len){        if(k==-1||p[k]==p[j]){            ++k;            ++j;            next[j]=k;        }        else k=next[k];    }    //for(i=0;i<=len;i++) printf("%d ",next[i]);    //printf("\n");}int main(){    int i,j,k;    int num;    int n;    scanf("%d",&num);    while(num--){        scanf("%d",&n);        scanf("%s",a);        getnext(a);        int len=strlen(a);        int sum=0;        for(i=1;i<=len;i++){            dt[i]=dt[next[i]]+1;            dt[i]%=10007;            sum+=dt[i];            sum%=10007;        }        printf("%d\n",sum);    }    return 0;}