D D - Om Nom and Necklace

官方题解（讲的非常详细）

This task is to determine whether a string is in the form of ABABA... ABA for each prefixes of a given string S

For a prefix P, let's split it into some blocks, just like P = SSSS... SSSST, which T is a prefix of S. Obviously, if we use KMP algorithm, we can do it in linear time, and the length of S will be minimal. There are only two cases : T = S, T ≠ S.

1. T = S. When T = S, P = SSS... S. Assume that S appears R times. Consider "ABABAB....ABABA", the last A must be a suffix of P, and it must be like SS... S, so A will be like SS... SS, and so will B. By greedy algorithm, the length of A will be minimal, so it will be SSS... S, where S appears times. And B will be SSS... S, where S appears times. So we just need to check whether .
2. T ≠ S . When T ≠ S, the strategy is similar to T = S. A will be like "SS...ST", and its length will be minimal. At last we just need to check whether .

The total time complexity is O(n).

AC代码

#include 
#include 
#include 
#include 
#include 

using namespace std;

const int MAX = 1000010;

int n, k;
char s[MAX];
int next[MAX];

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

int main(){
	scanf("%d%d", &n, &k);
	scanf("%s", s);
	make_next();
	for(int i=1; i<=n; i++){
		int cur = i - next[i];
		int L = i % cur;
		int cnt = i / cur;
		if(L  == 0){
			if(cnt / k - cnt % k >= 0) printf("1");
			else printf("0");
		}
		else{
			if(cnt / k - cnt % k > 0) printf("1");
			else printf("0");
		}
	}	
	puts("");
}