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  • Codeforces Round #628 (Div. 2) A. EhAb AnD gCd

    You are given a positive integer xx . Find any such 22 positive integers aa and bb such that GCD(a,b)+LCM(a,b)=xGCD(a,b)+LCM(a,b)=x .

    As a reminder, GCD(a,b)GCD(a,b) is the greatest integer that divides both aa and bb . Similarly, LCM(a,b)LCM(a,b) is the smallest integer such that both aa and bb divide it.

    It's guaranteed that the solution always exists. If there are several such pairs (a,b)(a,b) , you can output any of them.

    Input

    The first line contains a single integer tt (1t100)(1≤t≤100)  — the number of testcases.

    Each testcase consists of one line containing a single integer, xx (2x109)(2≤x≤109) .

    Output

    For each testcase, output a pair of positive integers aa and bb (1a,b109)1≤a,b≤109) such that GCD(a,b)+LCM(a,b)=xGCD(a,b)+LCM(a,b)=x . It's guaranteed that the solution always exists. If there are several such pairs (a,b)(a,b) , you can output any of them.

    Example
    Input
    Copy
    2
    2
    14
    
    Output
    Copy
    1 1
    6 4
    直接输出1和x-1即可。
    #include <bits/stdc++.h>
    using namespace std;
    int main()
    {
        int t;
        cin>>t;
        while(t--)
        {
            int x;
            cin>>x;
            cout<<1<<' '<<x-1<<endl;
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/lipoicyclic/p/12498890.html
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