Happy 2006
| Time Limit: 3000MS | Memory Limit: 65536K | |
| Total 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.
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
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 1000000
LL 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;
}