题意:
求出所有的前缀的周期循环次数大于1的,输出长度,循环次数
分析:
还是周期=i−nxt[i],然后搞就好了
代码:
//
// Created by TaoSama on 2015-10-28
// Copyright (c) 2015 TaoSama. All rights reserved.
//
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>
using namespace std;
#define pr(x) cout << #x << " = " << x << " "
#define prln(x) cout << #x << " = " << x << endl
const int N = 1e6 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;
int n, nxt[N];
char s[N];
void getNxt() {
nxt[0] = -1;
int i = 0, j = -1;
while(i < n) {
if(j == -1 || s[i] == s[j]) nxt[++i] = ++j;
else j = nxt[j];
}
}
int main() {
#ifdef LOCAL
freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);
// freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);
#endif
ios_base::sync_with_stdio(0);
int kase = 0;
while(scanf("%d", &n) == 1 && n) {
scanf("%s", s);
getNxt();
printf("Test case #%d\n", ++kase);
for(int i = 2; i <= n; ++i) {
int cycle = i - nxt[i];
int ans = i % cycle ? 0 : i / cycle > 1 ? i / cycle : 0;
if(ans) printf("%d %d\n", i, ans);
}
puts("");
}
return 0;
}