Happy 2006 欧几里得定理

Posted 2020-09-08 joeylee97

Happy 2006

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 11956		Accepted: 4224

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

Source

 

 

周期性！

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
typedef long long LL;
#define MAXN 1000002
/*
要求找到与m互质的第k大的数字
gcd(a,b) = gcd(b*t+a,b)
*/
LL a[MAXN];
LL gcd(LL a,LL b)
{
    if(b==0)
        return a;
    return gcd(b,a%b);
}
int main()
{
    LL m,k;
    while(scanf("%lld%lld",&m,&k)!=EOF)
    {
        LL p=0;
        for(int i=1;i<=m;i++)
        {
            if(gcd(i,m)==1)
                a[++p] = i;
        }
        if(k%p==0)
            printf("%lld\n",(k/p-1)*m+a[p]);
        else
            printf("%lld\n",(k/p)*m+a[k%p]);
    }
}