HUD 3336 Count the string(kmp+思维)

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分类专栏： HDU习题集字符串处理

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本文介绍了一个字符串处理问题：计算一个字符串的所有非空前缀在原字符串中出现的总次数，并提供了一个具体的算法实现，该算法利用了Next数组来高效解决这一问题。

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Count the string

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 12914 Accepted Submission(s): 5948

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

Author

foreverlin@HNU

Source

HDOJ Monthly Contest – 2010.03.06

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#include<iostream>
#include<algorithm>
#include<cstring>
#include<stdio.h>
using namespace std;
typedef long long LL;
#define maxn (200005)
#define MOD 10007
int n;
int Next[maxn],cnt;
char seq[maxn];
/*
题意是求每个前缀在原串中的出现次数之和。。。
想法很简单，在求Next数组时如果求出来的Next[i]不为-1，
也就是存在前缀后缀相等，则计数，
（题意貌似没有明确说明其不能重叠）
对于每一个前缀，如果出现过，一定可以通过这种方法找出来并且计数，
至于重复情况，因为i是向前扫的，所以不存在。。。

*/
void getNext(int len)
{
    cnt=0;
    //memset(cnt,0,sizeof(cnt));
    Next[0]=-1;
    int k=-1;
    for(int i=1;i<len;i++)
    {
        while(k>-1&&seq[k+1]!=seq[i])
            k=Next[k];
          //  cout<<seq[k+1]<<" "<<seq[i]<<endl;
        if(seq[k+1]==seq[i]) k++;
        Next[i]=k;
        //cout<<k<<endl;
        if(k>=0) cnt++;
    }
}
int main()
{
    int t;cin>>t;
    while(t--)
    {
        int i;
        scanf("%d",&n);
        for( i=0;i<n;i++) cin>>seq[i];
        seq[i]=0;
        int len=strlen(seq);
        getNext(len);
        printf("%d\n",(cnt+len)%MOD);
    }
    return 0;
}