2013 - ECJTU 暑期训练赛第七场-problem-G

C -O.O

Crawling in process...Crawling failedTime Limit:3000MS    Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

SubmitStatus Practice POJ 2773

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 12006 22006 3

Sample Output

135

AC代码+解释：

/*你求出了1~n与n互质的数之后，把这些数加上n就可以得到n~2n间的与n互质的数。而且n~2n间不会有某个与n互质的数被漏掉。因为如果n<=a<=2n，且gcd(a, n)==1，那么gcd(a - n, n)必然等于1。也就是必然有个在1~n间的数a-n，可以通过加n的方式得到a。所以与n互质的数是有周期性的。我们只需要求出第一个周期即可。*/#include<iostream>#include<string>#include<cstring>#include<iomanip>#include<algorithm>#include<cmath>#include<queue>#include<stack>#include<map>#include<cstdio>#include<cstdlib>const int MAX=1000001;int prim[MAX];int m;using namespace std;int GCD(int x,int y)//求公约数{    if(y==0)    return x;    else    return GCD(y,x%y);}int main(){    __int64 k;    int i,j,n,t;    prim[0]=1;    while(cin>>m>>k)    {        n=1;        for(i=2;i<=m;i++)        {            if(GCD(m,i)==1)            {                prim[n]=i;                n++;            }        }        if(k%n!=0)        {            t=k/n;            cout<<m*t+prim[k%n-1]<<endl;        }        else        {            t=k/n-1;            cout<<m*t+prim[n-1]<<endl;        }    }    return 0;}