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 << endl
const int N = 1e6 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;

int n, nxt
;
char s
;

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
; 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;
}