Happy 2006 素数打表+容斥原理+二分+hdu

Time Limit: 3000MS Memory Limit: 65536KB 64bit IO Format: %I64d & %I64u


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

解决方案:题意是求第k个与m互质的数是哪个数。假设m的质因数为p1,p2,p3.

那么若s与m互质,它将是第s-s/p1-s/p2-s/p3+s/(p1*p2)+s/(p1*p3)+s/(p2*p3)-s/(p1*p2*p3)个数。所以,可以通过二分查找来确定这个数。

code:

#include<iostream>
#include<cstdio>
#include<vector>
#include<cstring>
#define MMAX 1000000
using namespace std;
bool vis[MMAX];
vector <int > p,f;
void init()
{
    p.clear();
    memset(vis,true,sizeof(vis));
    p.push_back(2);
    for(int i=3; i<MMAX; i+=2)
    {
        if(vis[i])
        {
            p.push_back(i);
            for(int j=i+i; j<MMAX; j+=i)
                vis[j]=false;
        }
    }

}///素数打表

void prime()
{
    p.clear();
    memset(vis, true, sizeof(vis));
    p.push_back(2);
    for (int i=3; i<MMAX; i+=2)
        if (vis[i])
        {
            p.push_back(i);
            for (int j=i+i; j<MMAX; j+=i) vis[j] = false;
        }
}
long long ok(long long mid)
{
    long long c,v,kk,all=mid;
    for(int i=1; i<(1<<f.size()); i++)
    {
        c=0,v=1;
        for(int j=0; j<f.size(); j++)
        {
            if((1<<j)&i)
            {
                c++,v*=f[j];
            }
        }
        kk=mid/v;
        if(c%2==1) all-=kk;
        else all+=kk;
    }
    return all;
}
int main()
{
    long long m,k;
    //prime();
    init();
    //cout<<p[0];
    while(~scanf("%lld%lld",&m,&k))
    {
        f.clear();
        for(unsigned i=0; i<p.size()&&p[i]<=m; i++)
        {
            if(m%p[i]==0)
            {
                f.push_back(p[i]);
                while(m%p[i]==0) m/=p[i];
            }
        }
        long long  low=1,high=1e17,mid;
        while(low<high)
        {
            mid=(low+high)/2;
            if(ok(mid)>=k) high=mid;
            else low=mid+1;
        }
        cout<<low<<endl;
    }
    return 0;
}

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