HDU 1358 Period （kmp求周期）

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 << endlconst 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;}