HUST 1010 The Minimum Length（kmp求周期）

题意:

有一个字符串A，假设A是“abcdefg”,由A可以重复组成无线长度的AAAAAAA，即“abcdefgabcdefgabcdefg.....”.
从其中截取一段"abcdefgabcdefgabcdefgabcdefg",取红色部分为截取部分,设它为字符串B
现在先给出字符串B,求A最短的长度

分析:

说了这么多废话其实就是求最小循环节=n−nxt[n]

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);    while(scanf("%s", s) == 1) {        n = strlen(s);        getNxt();        printf("%d\n", n - nxt[n]);    }    return 0;}