POJ 2773 Happy 2006(二分+容斥）

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

分析：问你和n互斥的第k个数其实直接二分就行了

#include<cstdio>#include<cstring>#include<algorithm>#include<vector>#include<string>#include<iostream>#include<queue>#include<cmath>#include<map>#include<stack>#include<set>using namespace std;#define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i )#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )#define CLEAR( a , x ) memset ( a , x , sizeof a )const int INF=0x3f3f3f3f;typedef long long LL;LL n,k;int v[10],m;LL solve(LL x){    LL sum=0;    for(int i=1;i<(1<<m);i++)    {        LL c=0,s=1;        for(int j=0;j<m;j++)        {            if(i&(1<<j))            {                s=s*v[j];                c++;            }        }        if(c&1)            sum+=x/s;        else            sum-=x/s;    }    return x-sum;}int main(){    while(~scanf("%d%d",&n,&k))    {        m=0;        for(int i=2;i*i<=n;i++)        {            if(n%i==0)            {                while(n%i==0)                    n/=i;                v[m++]=i;            }        }        if(n>1) v[m++]=n;        LL l=1,r=1e18;        while(l<=r)        {            LL mid=(l+r)>>1;            if(solve(mid)>=k)                r=mid-1;            else                l=mid+1;        }        printf("%lld\n",l);    }    return 0;}