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H - Happy 2006

2024-07-31 12:52:42   220人阅读

H -Happy 2006

Time Limit:3000MS    Memory Limit:65536KB    64bit IO Format:%I64d & %I64u
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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
 意解:题目简单易懂,但是要解决这道题目就需要费点功夫了.首先要知道如果一个数与另一个数的gcd不为1,则它们有公共的质因子,所以首先要先算出n的所有最小质因子,然后分块,加容斥就可以了. 怎么分块呢? 可以这样理解,对于一段数的区间{l,r},可以知道有多少个跟n不互质,这个可以用容斥原理搞定,剩下的就是不断延长区间.形如[l + x,r + y],这种形式;具体看代码吧!AC代码:
#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>

using namespace std;
typedef long long ll;

vector<int>tp;

int main()
{
    int n,k,t;
    while(~scanf("%d %d",&n,&k))
    {
        tp.clear();
        t = n;
        for(int i = 2; i * i <= n; i++)
        {
            if(t % i == 0)
            {
                tp.push_back(i);
                while(!(t % i))
                    t /= i;
            }
        }
        if(t != 1) tp.push_back(t);
        ll now,next,Len,ans;
        Len = tp.size();
        now = 1;
        next = now + k - 1;
        while(k)
        {
            ans = 0;
            for(int i = 1; i < (1LL << Len); i++)
            {
                int mul = 1,tmp = 0;
                for(int j = 0; j < Len; j++)
                    if((i >> j & 1)) mul *= tp[j],tmp++;
                if(tmp & 1) ans += next / mul - (now - 1) / mul;
                else ans -= next / mul - (now - 1) / mul;
            }
            k = k - (next - now + 1 - ans);
            if(!k) break;
            now = next + 1;
            next = now + k - 1;
        }
        printf("%I64d\n",next);
    }
    return 0;
}



H - Happy 2006


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