H - Happy 2006

標籤：數學   容斥   

H -Happy 2006

Time Limit:3000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64uSubmit   Status

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 12006 22006 3

Sample Output

135 意解:題目簡單易懂,但是要解決這道題目就需要費點功夫了.首先要知道如果一個數與另一個數的gcd不為1,則它們有公用的質因子,所以首先要先算出n的所有最小質因子,然後分塊,加容斥就可以了. 怎麼分塊呢? 可以這樣理解,對於一段數的區間{l,r},可以知道有多少個跟n不互質,這個可以用容斥原理搞定,剩下的就是不斷延長區間.形如[l + x,r + y],這種形式;具體看代碼吧!AC代碼:
#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