POJ 2773 Happy 2006

Happy 2006

Time Limit: 3000ms

Memory Limit: 65536KB

This problem will be judged on PKU. Original ID: 2773
64-bit integer IO format: %lld      Java class name: Main

 

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

 

解题：容斥原理

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
typedef long long LL;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int maxn = 50;
int p[maxn],tot;
void init(int x){
    tot = 0;
    for(int i = 2; i*i <= x; ++i){
        if(x%i == 0){
            p[tot++] = i;
            while(x%i == 0) x /= i;
        }
    }
    if(x > 1) p[tot++] = x;
}
LL solve(LL x){
    LL ret = 0;
    for(int i = 1; i < (1<<tot); ++i){
        int cnt = 0,tmp = 1;
        for(int j = 0; j < tot; ++j){
            if((i>>j)&1){
                ++cnt;
                tmp *= p[j];
            }
        }
        if(cnt&1) ret -= x/tmp;
        else ret += x/tmp;
    }
    return x + ret;
}
int main(){
    int n,k;
    while(~scanf("%d%d",&n,&k)){
        LL ret = 0,low = 1,high = INF;
        init(n);
        while(low <= high){
            LL mid = (low + high)>>1;
            if(solve(mid) >= k){
                ret = mid;
                high = mid - 1;
            }else low = mid + 1;
        }
        printf("%I64d
",ret);
    }
    return 0;
}