Mean:
給你一個字元串,讓你從第二個字元開始判斷目前長度的字元串是否是重複串,如果是,輸出目前位置,并輸出重複串的周期。
analyse:
還是next數組的運用,詳見上一篇部落格。
Time complexity: O(N)
Source code:
/*
* this code is made by crazyacking
* Verdict: Accepted
* Submission Date: 2015-07-28-07.12
* Time: 0MS
* Memory: 137KB
*/
#include <queue>
#include <cstdio>
#include <set>
#include <string>
#include <stack>
#include <cmath>
#include <climits>
#include <map>
#include <cstdlib>
#include <iostream>
#include <vector>
#include <algorithm>
#include <cstring>
#define LL long long
#define ULL unsigned long long
using namespace std;
const int MAXN=1000010;
int n;
char s[MAXN];
int Next[MAXN];
void getNext()
{
Next[0]=0;
for(int i=1,k=0;i<n;++i)
{
while(s[i]!=s[k]&&k) k=Next[k-1];
if(s[i]==s[k]) ++k;
Next[i]=k;
}
}
int main()
ios_base::sync_with_stdio(false);
cin.tie(0);
int cas=1;
while(~scanf("%d",&n)&&n)
scanf("%s",s);
getNext();
printf("Test case #%d\n",cas++);
for(int i=1;i<n;++i)
{
int now_cycle=(i+1)-Next[i];
if((now_cycle!=i+1) && (i+1)%now_cycle==0)
{
printf("%d %d\n",i+1,(i+1)/now_cycle);
}
}
puts("");
return 0;