POJ 2773 Happy 2006

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

Source

POJ Monthly--2006.03.26,static

题目大意：

给定m,k，问你第K个与m互质的数是多少？  其中  m (1 <= m <= 1000000), K (1 <= K <= 100000000).

解题思路：

用位运算的容斥原理，计算 [1，x]与m互质的数的方法是：

假设 m的质因数为 a,b,c ，那么与m不互斥的数个数为 f(a)+f(b)+f(c)-f(ab)-f(ac)-fa(bc)+f(abc)，f(t)的含义是 (1,x)有多少个数与t不互质，很明显f(t)=x/t，那么与m互斥的数个数很明显为x-( f(a)+f(b)+f(c)-f(ab)-f(ac)-fa(bc)+f(abc)).

因此只要二分求出刚好大于等于K的这个数就是答案。

解题代码：

#include <iostream>
#include <cstdio>
#include <vector>
using namespace std;

typedef long long ll;
const int maxn=1100000;
int m,n;
vector <int> v;
vector <int> t;
bool f[maxn];

void getPrime(){
    for(int i=0;i<maxn;i++) f[i]=true;
    for(int i=4;i<maxn;i+=2) f[i]=false;
    for(int i=3;i<maxn;i+=2){
        for(int j=i;j<maxn/i;j+=2){
            f[i*j]=false;
        }
    }
    for(int i=2;i<maxn;i++){
        if(f[i]) v.push_back(i);
    }
}

void getsubPrime(){
    t.clear();
    int s=0,now=m;
    while(now>=v[s]){
        if(now%v[s]==0){
            t.push_back(v[s]);
            while(now%v[s]==0){
                now/=v[s];
            }
        }
        s++;
    }
}

ll get(ll y){
    ll ans=0;
    for(int i=0;i<(1<<t.size());i++){
        int cnt=0,sum=1;
        for(int k=0;k<t.size();k++){
            if( i& (1<<k) ){
                sum*=t[k];
                cnt++;
            }
        }
        if(cnt%2==0) ans+=y/sum;
        else ans-=y/sum;
    }
    return ans;
}

void solve(){
    getsubPrime();
    ll l=0,r=1e17;
    while(l<r){
        ll mid=(l+r)/2;
        if(get(mid)>=n) r=mid;
        else l=mid+1;
    }
    cout<<r<<endl;
}

int main(){
    getPrime();
    while(scanf("%d%d",&m,&n)!=EOF){
        solve();
    }
    return 0;
}

POJ 2773 Happy 2006