POJ 2773 Happy 2006（容斥原理+二分）

Happy 2006

Time Limit: 3000MS Memory Limit: 65536KTotal Submissions: 10827 Accepted: 3764

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

这道题目是HDU 3388的简化版，方法几乎一模一样

http://blog.csdn.net/dacc123/article/details/51285731

#include <iostream>#include <string.h>#include <stdlib.h>#include <algorithm>#include <math.h>#include <stdio.h>using namespace std;typedef long long int LL;const LL INF=(LL)1<<62;#define MAX 1000000LL prime[MAX+5];LL sprime[MAX+5];LL q[MAX+5];LL check[MAX+5];LL m,k,cnt;void eular(){    memset(check,false,sizeof(check));    int tot=0;    for(int i=2;i<=MAX+5;i++)    {        if(!check[i])            prime[tot++]=i;        for(int j=0;j<tot;j++)        {            if(i*prime[j]>MAX+5) break;            check[i*prime[j]]=true;            if(i%prime[j]==0) break;        }    }}void Divide(LL n){    cnt=0;    LL t=(LL)sqrt(1.0*n);    for(LL i=0;prime[i]<=t;i++)    {        if(n%prime[i]==0)        {            sprime[cnt++]=prime[i];            while(n%prime[i]==0)                n/=prime[i];        }    }    if(n>1)        sprime[cnt++]=n;}LL Ex(LL n){    LL sum=0,t=1;    q[0]=-1;    for(LL i=0;i<cnt;i++)    {         LL x=t;        for(LL j=0;j<x;j++)        {            q[t]=q[j]*sprime[i]*(-1);            t++;        }    }    for(LL i=1;i<t;i++)        sum+=n/q[i];    return sum;}LL binary(){    LL l=1,r=INF;    LL mid,ans;    while(l<=r)    {        mid=(l+r)/2;        if((mid-Ex(mid))>=k)        {            r=mid-1;        }        else            l=mid+1;    }    return l;}int main(){    eular();    while(scanf("%lld%lld",&m,&k)!=EOF)    {        if(m==1)        {            printf("%lld\n",k);            continue;        }        Divide(m);        printf("%lld\n",binary());    }    return 0;}