poj2773 Happy 2006

最新推荐文章于 2019-01-31 10:52:39 发布

原创最新推荐文章于 2019-01-31 10:52:39 发布 · 831 阅读

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本文讨论了如何通过二分枚举和容斥原理找到给定整数m的第k个相对质数，详细解释了解题思路和算法实现。

Happy 2006

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 9987		Accepted: 3434

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

解题思路：二分枚举1-2^64之间的数x，找到1-x与m互质的个数s，这里利用容斥原理，如果s与k相等，那么该x即为结果；

参考代码：

#include <iostream>
#include <vector>
#include <math.h>
using namespace std;
typedef long long ll;
const int MAX = 1000010;
int p[MAX],count;
void fun(ll n){
	ll i;
	count=0;
	for (i=2;i*i<=n;i++){
		if (n%i==0){
			p[count++]=i;
			while (n%i==0)
				n/=i;
		}
	}
	if (n>1)
		p[count++]=n;
}
ll solve(ll n,ll r){
	ll sum=0;
	for (int i=1;i<(1<<count);i++){
		ll mult=1,bits=0;
		for (int j=0;j<count;j++){
			if (i&(1<<j)){
				bits++;
				mult*=p[j];
			}
		}
		ll cur=r/mult;
		if (bits%2==1)
			sum+=cur;
		else
			sum-=cur;
	}
	return r - sum;
}
int main(){
	ll n,m,mid,num;
	while (cin>>n>>m){
		fun(n);
		ll r=0xffffffff,l=1;
		while (r-l>0){
			mid=(r+l)/2;
			num=solve(n,mid);
			if (num>=m)
				r=mid;
			else
				l=mid+1;
		}
		cout<<l<<endl;
	}
	return 0;
}