POJ - 1961:Period

Period
出典:POJ
ラベル:KMPアルゴリズム
参考資料:
類似タイトル:
タイトル
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.
入力
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.
しゅつりょく
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.
入力サンプル
3 aaa 12 aabaabaabaab 0
出力サンプル
Test case #1 2 2 3 3 Test case #2 2 2 6 2 9 3 12 4
問題を解く構想.
KMPアルゴリズムでnext配列を求める方法を参照してください.注意、この問題は食べる時間が非常にきついので、cinとcoutの代わりにscanfとprintを使用し、strlen()を毎回呼び出して文字列の長さを求めることを避けます.
リファレンスコード

#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
", ++kase);
		for(int i=1; i<=n; ++i){
			int period=i-next[i];
			if(i!=period && i%period==0)
				printf("%d %d
", i, i/period);
		}
		printf("
");
	}
}