POJ - 1961：Period

Period

来源：POJ

标签：KMP算法

参考资料：

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题目

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.

输入

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.

输出

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.

输入样例

3
aaa
12
aabaabaabaab
0

输出样例

Test case #1
2 2
3 3


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

解题思路

参考KMP算法中求next数组的方法。注意，这道题吃时间非常紧，使用scanf和print代替cin和cout，同时避免在每次循环中调用strlen()来求字符串长度。

参考代码

#include
#include
#define N 1000005

char str[N];
int n;
int next[N];

void getNext(char *T){
	int i=0, j=-1;
	next[0]=-1;
	while(i<n){
		if(j==-1 || T[i]==T[j]){
			++i; ++j;
			next[i]=j;
		}
		else j=next[j];
	}
}


int main(){
	int kase=0;
	while(scanf("%d", &n) && n!=0){
		scanf("%s", str);
		getNext(str);
		printf("Test case #%d\n", ++kase);
		for(int i=1; i<=n; ++i){
			int period=i-next[i];
			if(i!=period && i%period==0)
				printf("%d %d\n", i, i/period);
		}
		printf("\n");
	}
}