Codeforces Round #275 (Div. 2) --A Countexample

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最新推荐文章于 2024-08-02 17:51:14 发布

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分类专栏：暴力 Codeforces

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本文探讨了一道关于互素数的问题，旨在找到三个特定整数a、b、c，使得(a, b)与(b, c)互素，而(a, c)不互素，并给出了相应的解决思路及代码实现。

题目链接：Counterexample

Your friend has recently learned about coprime numbers. A pair of numbers {a, b} is called coprime if the maximum number that divides both a and b is equal to one.

Your friend often comes up with different statements. He has recently supposed that if the pair (a, b) is coprime and the pair (b, c) is coprime, then the pair (a, c) is coprime.

You want to find a counterexample for your friend's statement. Therefore, your task is to find three distinct numbers (a, b, c), for which the statement is false, and the numbers meet the condition l ≤ a < b < c ≤ r.

More specifically, you need to find three numbers (a, b, c), such that l ≤ a < b < c ≤ r, pairs (a, b) and (b, c) are coprime, and pair (a, c)is not coprime.

Input

The single line contains two positive space-separated integers l, r (1 ≤ l ≤ r ≤ 1018; r - l ≤ 50).

Output

Print three positive space-separated integers a, b, c — three distinct numbers (a, b, c) that form the counterexample. If there are several solutions, you are allowed to print any of them. The numbers must be printed in ascending order.

If the counterexample does not exist, print the single number -1.

   Sample test(s)
 
     input
   
2 4

     output
   
2 3 4

     input
   
10 11

     output
   
-1

     input
   
900000000000000009 900000000000000029

     output
   
900000000000000009 900000000000000010 900000000000000021

Note

In the first sample pair (2, 4) is not coprime and pairs (2, 3) and (3, 4) are.

In the second sample you cannot form a group of three distinct integers, so the answer is -1.

In the third sample it is easy to see that numbers 900000000000000009 and 900000000000000021 are divisible by three.

大致题意：给定l和r，找出满足条件 l ≤ a < b < c ≤ r 的a, b, c，且a和b互素，b和c互素，但a和c不互素。

解题思路：暴力枚举。枚举所有的情况，对每个情况进行判断。这种做法的前提是数据不大，数据要是大了，就不能这么直接暴力了。

AC代码：

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;
#define INF 0x7fffffff
#define LL long long

LL gcd(LL a, LL b){
    if(!b) return a;
    return gcd(b, a%b);
}

int main()
{
    #ifdef sxk
        freopen("in.txt","r",stdin);
    #endif
    LL l,r,a,b,c;
    while(scanf("%lld%lld",&l,&r)!=EOF)
    {
        int flag = 0;
        for(LL i=l; i<=r; i++){
            for(LL j=i+1; j<=r; j++){
                for(LL k=j+1; k<=r; k++){
                    if(gcd(i, j)==1 && gcd(j, k)==1 && gcd(i, k)!=1)
                    {
                        printf("%lld %lld %lld\n", i, j, k);
                        flag = 1;
                        break;
                    }
                }
                if(flag)  break;
            }
             if(flag)  break;
        }
        if(!flag)  printf("%d\n", -1);
    }
    return 0;
}