POJ 1961 Period

Period

Time Limit: 3000ms

Memory Limit: 30000KB

This problem will be judged on PKU. Original ID: 1961
64-bit integer IO format: %lld      Java class name: Main

 

For each prefix of a given string S with N characters (each character has an ASCII code between 97 and 126, inclusive), we want to know whether the prefix is a periodic string. That is, for each i (2 <= i <= N) we want to know the largest K > 1 (if there is one) such that the prefix of S with length i can be written as AK ,that is A concatenated K times, for some string A. Of course, we also want to know the period K.

 

Input

The input consists of several test cases. Each test case consists of two lines. The first one contains N (2 <= N <= 1 000 000) – the size of the string S.The second line contains the string S. The input file ends with a line, having the 
number zero on it.

 

Output

For each test case, output "Test case #" and the consecutive test case number on a single line; then, for each prefix with length i that has a period K > 1, output the prefix size i and the period K separated by a single space; the prefix sizes must be in increasing order. Print a blank line after each test case.

 

Sample Input

3
aaa
12
aabaabaabaab
0

Sample Output

Test case #1
2 2
3 3

Test case #2
2 2
6 2
9 3
12 4

Source

Southeastern Europe 2004

 

解题：KMP。求串前缀中是周期串的信息，输出前缀串长，以及周期数。

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <climits>
#include <vector>
#include <queue>
#include <cstdlib>
#include <string>
#include <set>
#include <stack>
#define LL long long
#define pii pair<int,int>
#define INF 0x3f3f3f3f
using namespace std;
const int maxn = 1000010;
char str[maxn];
int n,fail[maxn],cs = 1;
void getFail(){
    fail[0] = fail[1] = 0;
    int i;
    printf("Test case #%d
",cs++);
    for(i = 1; str[i]; i++){
        int j = fail[i];
        if(i%(i-fail[i]) == 0 && i/(i - fail[i]) > 1)
            printf("%d %d
",i,i/(i-fail[i]));
        while(j && str[i] != str[j]) j = fail[j];
        fail[i+1] = str[i] == str[j] ? j+1:0;
    }
    if(i%(i - fail[i]) == 0 && i/(i - fail[i]) > 1)
        printf("%d %d
",i,i/(i - fail[i]));
}
int main() {
    bool ok = false;
    while(scanf("%d",&n),n){
        if(ok) puts("");
        ok = true;
        scanf("%s",str);
        getFail();
    }
    return 0;
}