Count the string kmp next数组应用

最新推荐文章于 2024-07-27 14:57:47 发布

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最新推荐文章于 2024-07-27 14:57:47 发布

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分类专栏：字符串

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本文介绍了一个字符串处理问题，目标是统计一个给定字符串的所有非空前缀在整个字符串中出现的次数，并通过模运算处理可能的大数值结果。文章提供了一种有效的解决方案及AC代码实现。

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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

Hint

题意

统计每个前缀在整个串中出现次数

题解：

设数组a[i]表示以st[i]结尾的前缀出现次数
那么有递推式 a[i] = a[nxt[i]]+1; 即长度小于i以st[i]结尾的前缀数+该子串自己（1）

AC代码

#include <cstdio>
#include <cstring>
char st[200005];
int a[200005];
int nxt[200005];
void getnext(char *s,int l){
    int i = 0,j = 0;
    nxt[0] = -1; j = nxt[i];
    while (i<l){
        if (j == -1 || s[i]==s[j]){
            nxt[++i] = ++j;
        }else j = nxt[j];
    }
}
int main(){
    int t;
    scanf("%d",&t);
    while (t--){
        int len;
        scanf("%d",&len);
        scanf("%s",st);
        memset(a,0,sizeof(a));
        memset(nxt,0,sizeof(nxt));
        getnext(st,len);
        for (int i = 1;i <= len; ++i){
            a[i] = a[nxt[i]]+1;
        }
        int sum = 0;
        for (int i = 1;i <= len; ++i){
            sum = (sum+a[i])%10007;
        }
        printf("%d\n",sum);
    }
    return 0;
}