UVALive

Problem Description

Dr. Ellie Arroway has established contact with an extraterrestrial civilization. However, all efforts
to decode their messages have failed so far because, as luck would have it, they have stumbled upon
a race of stuttering aliens! Her team has found out that, in every long enough message, the most
important words appear repeated a certain number of times as a sequence of consecutive characters,
even in the middle of other words. Furthermore, sometimes they use contractions in an obscure manner.
For example, if they need to say bab twice, they might just send the message babab, which has been
abbreviated because the second b of the first word can be reused as the first b of the second one.
Thus, the message contains possibly overlapping repetitions of the same words over and over again.
As a result, Ellie turns to you, S.R. Hadden, for help in identifying the gist of the message.
Given an integer m, and a string s, representing the message, your task is to find the longest
substring of s that appears at least m times. For example, in the message baaaababababbababbab, the
length-5 word babab is contained 3 times, namely at positions 5, 7 and 12 (where indices start at zero).
No substring appearing 3 or more times is longer (see the first example from the sample input). On
the other hand, no substring appears 11 times or more (see example 2).
In case there are several solutions, the substring with the rightmost occurrence is preferred (see
example 3).

Input

The input contains several test cases. Each test case consists of a line with an integer m (m ≥ 1),
the minimum number of repetitions, followed by a line containing a string s of length between m and
40 000, inclusive. All characters in s are lowercase characters from ‘a’ to ‘z’. The last test case is
denoted by m = 0 and must not be processed.

Output

Print one line of output for each test case. If there is no solution, output ‘none’; otherwise, print two
integers in a line, separated by a space. The first integer denotes the maximum length of a substring
appearing at least m times; the second integer gives the rightmost possible starting position of such a
substring.

Sample Input

3
baaaababababbababbab
11
baaaababababbababbab
3
cccccc
0

Sample Output

5 12
none
4 2

题意：给定一个字符串，找出重复出现超过m次的字串的最大开始下标
思路： Hash + 二分 对字符串的长度进行二分 时间复杂度 O( n*lgn )

AC code：

/*给定一个字符串，找出重复出现超过m次的字串的最大开始下标*/
#include <map>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

typedef unsigned long long llu;

const int maxn = 4e4+50;

int len ,t ;
llu seed = 131,Hash[maxn],temp[maxn];
char str[maxn];

void init() {
    Hash[0] = 1;
    for (int i = 1;i <= len;i++) {
        Hash[i] = Hash[i-1]*seed + (str[i]-'a'+1);
    }
}

llu get(int l,int r) {
    return Hash[r] - Hash[l-1] * temp[r-l+1] ;
}

bool judge (int length) {
    map<llu,int>mp;
    for (int i = length;i <= len ; i ++ ) {
        if( ++mp[get(i-length+1,i)] >= t ) {
            return true;
        }
    }
    return false;
}

int main(){
    temp[0] = 1;
    for (int i = 1;i<=maxn;i++) {
        temp[i] = temp[i-1] * seed;
    }
    while ( ~scanf("%d",&t) && t ) {
        scanf("%s",str+1);
        len = strlen(str+1);
        init();
        int l = 1 ,r = len ,mid ;
        if ( !judge(l) ) { printf("none
");  continue; }
        while ( l <= r ) {
            mid = (l + r)>>1;
            if ( judge(mid) ) {
                l = mid + 1;
            } else {
                r = mid - 1;
            }
        }
        int ans;
        map<llu,int>mp;
        for (int i = r ;i <= len;i++) {
        //printf("%d %llu
",i,get(i-r+1,i));
            if ( ++mp[get(i-r+1,i)] >= t ) {
                ans = i-r;
            }
        }
        printf("%d %d
",r,ans);
    }
    return 0;
}
/*
3
baaaababababbababbab
11
baaaababababbababbab
3
cccccc
0
*/