Happy 2006

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
思路：由欧几里得gcd(a,b)=gcd(a+b*t,b)，枚举小于m的所有互素的数，然后这些数加上一个m也与m互素，可以由这样的周期性找到第k个的答案。

如果a与b互素，则b×t+a与b也一定互素，如果a与b不互素，则b×t+a与b也一定不互素
故与m互素的数对m取模具有周期性，则根据这个方法我们就可以很快的求出第k个与m互素的数
假设小于m的数且与m互素的数有k个，其中第i个是ai，则第m×k+i与m互素的数是k×m+ai

# include <cstdio># include <iostream>using namespace std;typedef long long ll;ll gcd (ll a,ll b)//欧几里得算法{    return b ? gcd(b,a%b) : a;}ll n[1000010],p,m,i,j;ll k,a,ans,pos;int main (){    while (cin >> m >> k)    {        p = 0;        for (i=1; i<=m; ++i)            if (gcd(i,m) == 1)            n[++p] = i;        a = k / p;        pos = k % p;        if (!pos)        {            pos = p;            a--;        }        ans = a*m + n[pos];        cout << ans << endl;    }    return 0;}