数论（lcm）

CodeForces - 1154G

You are given an array a consisting of n integers a1,a2,…,an

.

Your problem is to find such pair of indices i,j

(1≤i<j≤n) that lcm(ai,aj)

is minimum possible.

lcm(x,y)

is the least common multiple of x and y (minimum positive number such that both x and y

are divisors of this number).

Input

The first line of the input contains one integer n

(2≤n≤106) — the number of elements in a

.

The second line of the input contains n

integers a1,a2,…,an (1≤ai≤107), where ai is the i-th element of a

.

Output

Print two integers i

and j (1≤i<j≤n) such that the value of lcm(ai,aj) is minimum among all valid pairs i,j

. If there are multiple answers, you can print any.

Examples

Input

5
2 4 8 3 6

Output

1 2

Input

5
5 2 11 3 7

Output

2 4

Input

6
2 5 10 1 10 2

Output

1 4题意：求最小的lcm（a,b）的下标；

#include <bits/stdc++.h>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <cstdio>
#include <string>
#include <stdio.h>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <string.h>
#include <vector>
#define ME(x , y) memset(x , y , sizeof(x))
#define SF(n) scanf("%d" , &n)
#define rep(i , n) for(int i = 0 ; i < n ; i ++)
#define INF  0x3f3f3f3f3f3f3f3fLL
#define mod 1000000007
#define PI acos(-1)
using namespace std;
typedef long long ll ;
const int N = 1e7 + 5 ;
int p[N] , n;

int main()
{
    int x , flaga , flagb ;//值，左下标，右下标
    ll minv ;//最小lcm值
    while(~scanf("%d" , &n))
    {
        memset(p , 0 , sizeof(p));//标记数组
        minv = INF ;
        for(int i = 1 ; i <= n ; i++)
        {
            scanf("%d" , &x);
            if(p[x] && x < minv)//出现了两次且小于原lcm，更新
            {
                minv = x ;
                flaga = p[x];
                flagb = i ;
            }
            p[x] = i ;//标记该数出现
        }
        for(int i = 1 ; i < N && i < minv ; i++)//遍历1-1e7+5数
        {
            ll v = 0 ;
            int pos ;
            for(int j = i ; j < N && j < minv ; j+=i)//成倍数的遍历。
            {
                if(p[j])//该数出现
                {
                    if(v==0)//第一个数
                    {
                        v = j ;
                        pos = p[j];
                    }
                    else if(v / i * j < minv)//第二个数出现，且小于原lcm更新。
                    {
                        minv = v / i * j ;
                        flaga = pos ;
                        flagb = p[j];
                    }
                    else
                    {
                        break ;
                    }
                }
            }
        }
        if(flaga > flagb) swap(flaga , flagb);
        cout << flaga << " " << flagb << endl;
    }

    return 0 ;
}

原文地址：https://www.cnblogs.com/nonames/p/11821002.html