Period POJ - 1961 （KMP扩展—next数组应用）

最新推荐文章于 2020-07-16 09:14:20 发布

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本文介绍了一个字符串处理问题，即如何判断一个字符串的前缀是否由相同子串重复组成，并给出了解决方案。通过构建next数组来高效地解决该问题。

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Period POJ - 1961

For each prefix of a given string S with N characters (each character has an ASCII code between 97 and 126, inclusive), we want to know whether the prefix is a periodic string. That is, for each i (2 <= i <= N) we want to know the largest K > 1 (if there is one) such that the prefix of S with length i can be written as A K,that is A concatenated K times, for some string A. Of course, we also want to know the period K.

Input

The input consists of several test cases. Each test case consists of two lines. The first one contains N (2 <= N <= 1 000 000) – the size of the string S.The second line contains the string S. The input file ends with a line, having the
number zero on it.

Output

For each test case, output "Test case #" and the consecutive test case number on a single line; then, for each prefix with length i that has a period K > 1, output the prefix size i and the period K separated by a single space; the prefix sizes must be in increasing order. Print a blank line after each test case.

Sample Input

3
aaa
12
aabaabaabaab
0

Sample Output

Test case #1
2 2
3 3

Test case #2
2 2
6 2
9 3
12 4

题意：已知长度为n的字符串，让求i位置之前的字符串是由多少个相同的字符串拼接而成的，例如aabaabaabaab第2个位置之前为aa，是由两个a拼接而成，第6个位置之前是aabaab是由两个相同的aab组成。输出i以及相同的字串数

思路：同poj2406，next数组的扩展应用

AC代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<cmath>
#include<utility>
#include<set>
#include<vector>
#include<map>
#include<queue>
#include<stack>
#define maxn1 1000005
#define maxn2 10005
#define INF 0x3f3f3f3f
#define LL long long
#define ULL unsigned long long
#define E 1e-8
#define mod 1000000007
#define P pair<int,int>
using namespace std;
 
int n;
char str[1000001];
int nexts[1000001];
void GetNexts(){
    int j=0,k=-1;
    nexts[0]=-1;
    while(j<n){
        if(k==-1||str[j]==str[k]){
            j++;
            k++;
            nexts[j]=k;
        }
        else k=nexts[k];
    }
}
int main(){
    int T = 0;
    while(scanf("%d",&n)!=EOF&&n){
        scanf("%s",str);
        GetNexts();
        printf("Test case #%d\n",++T);
        for(int i=1;i<=n;++i){
            int len = i-nexts[i];
            if(nexts[i]>0&&i%len==0)
                printf("%d %d\n",i,i/len);
        }
        printf("\n");
    }
    return 0;
}