poj 2773 Happy2006【容斥原理】

Happy 2006

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

题目翻译：给出m,k，求第k个和m互质的数字！

解题思路：数据量较大，因此全部采用long long，想求出第k个互质的数，数据量太大，因此采用2分查找，来找d第K个数字，想知道当前数字t是第几个与m的互质的数，需要求出t前有几个与m不互质的数字，利用容斥原理，可以解决，然后对比即可！

#include<cstdio>
#define LL long long
LL p[20],top;
void getp(LL m){
    LL i;
    for(i=2,top=0;i*i<=m;i++)
        if(m%i==0){
            p[top++]=i;
            while(m%i==0) m/=i;
        }
    if(m>1) p[top++]=m;
}
LL nop(LL mid,LL t){
    LL i,sum=0;
    for(i=t;i<top;i++)
        sum+=mid/p[i]-nop(mid/p[i],i+1);
    return sum;
}
int main(){
    LL k,m;
    while(scanf("%lld%lld",&m,&k)==2){
        LL mid,l=0,r=0x3f3f3f3f3f3f3f3f,t,ans=0;
        getp(m);
        while(l<=r){
            mid=(l+r)>>1;
            t=mid-nop(mid,0);
            if(t>=k){
                r=mid-1;
                if(t==k) ans=mid;
            }
            else l=mid+1;
        }
        printf("%lld\n",ans);
    }
    return 0;
}