FZU 1901 Period II （kmp）

题意:

L<=106字符串，对于每个长度为p的前缀，如果s[i]==s[p+i](p+i<L)，i∈[0,p)
那么我们认为它是一个periodic prefixs.求所有满足题意的前缀的长度p

分析:

就是求所有相同的前缀和后缀的位置,本来求这个东西扫一遍nxt数组就好了
现在要求位置,观察可以发现位置其实就是i−nxt[i]

代码:

////  Created by TaoSama on 2015-10-30//  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 t; scanf("%d", &t);    int kase = 0;    while(t--) {        scanf("%s", s);        n = strlen(s);        getNxt();        vector<int> ans;        for(int i = nxt[n]; i; i = nxt[i]) ans.push_back(n - i);        ans.push_back(n);        printf("Case #%d: %d\n", ++kase, ans.size());        for(int i = 0; i < ans.size(); ++i)            printf("%d%c", ans[i], " \n"[i == ans.size() - 1]);    }    return 0;}