Problem Description

Two positive integers are said to be relatively prime to each other if the Great Common Divisor (GCD) is 1. For instance, 1, 3, 5, 7, 9...are all relatively prime to 2006.

Now your job is easy: for the given integer m, find the K-th element which is relatively prime to m when these elements are sorted in ascending order.

Input 

The input contains multiple test cases. For each test case, it contains two integers m (1 <= m <= 1000000), K (1 <= K <= 100000000).

Output

Output the K-th element in a single line.

Sample Input

2006 1
2006 2
2006 3

Sample Output

1
3
5 

————————————————————————————————————————————————————

题意：给出 n、k 两个数，从 2 开始，求第 k 个 n 互素的数

思路：

没思路。。。搜了几篇题解，发现了这么一条性质：如果 a 与 b 互素，那么 b*t+a 肯定与 b 也互素

即：GCD(b*t+a,b)=GCD(a,b)

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因此，与 n 互素的数对 n 取模具有周期性，根据这个方法可以很快的求出第 k 个与 n 互素的数

假设小于 n 的数且与 n 互素的数有 k 个，第 i 个是 ai，则第 n*k+i 与 n 互素的数是 k×n+ai

Source Program

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<string>
#include<cstdlib>
#include<queue>
#include<set>
#include<map>
#include<stack>
#include<ctime>
#include<vector>
#define INF 0x3f3f3f3f
#define PI acos(-1.0)
#define N 1000005
#define MOD 1e9+7
#define E 1e-6
#define LL long long
using namespace std;
int a[N];
int GCD(int x,int y)
{
    if(y==0)
        return x;
    else
        return GCD(y,x%y);
}
int main()
{
    int n,k;
    while(scanf("%d%d",&n,&k)!=EOF)
    {
        int j=0;
        for(int i=1;i<=n;i++)
            if(GCD(n,i)==1)
                a[j++]=i;

        if(k%j)
            printf("%d\n",k/j*n+a[k%j-1]);
        else
            printf("%d\n",(k/j-1)*n+a[j-1]);
    }

    return 0;
}