poj2773 欧拉函数+整除性质

Happy 2006

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 10751		Accepted: 3744

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

题意：求出与m互素的第k大的数
首先，介绍一个定理：若a与b不互素,则,tb+a与b也不互素，若a与b互素，则tb+a与b互素。所以，这个题目可以先求出m以内的与m互素的数，大于m的可以用l*m+a求出，其中a为小于m的与m互素的数，l为循环次数。
用欧拉函数求出循环长度，同时对m以内的素数（即大于m的素数%m的余数）进行标记
（吐槽一下，昨天用sublime打欧拉的模板打错了，在dev上改了过来，sublime上保存的依然是错的，今天拿着错误的模板wa了好多发……也是醉了）

#include <stdio.h>
#include<string.h>
long long m,k;
int pan[1000008];
long long phi(long long n)
{ long long m=n;
long long i,cnt,j;
cnt=n;
for(i=2;i*i<=n;i++)
{
if(n%i==0)
{
for(j=i;j<=m;j=j+i)pan[j]=1;
cnt=cnt-cnt/i;
while(n%i==0)n=n/i;

}
}
//printf("n:%lld\n",n);
if(n>1){for(j=n;j<=m;j=j+n)pan[j]=1;cnt=cnt-cnt/n;}
return cnt;
}

int main(int argc, char const *argv[])
{ long long i,n;
while(~scanf("%lld%lld",&m,&k))
{ if(m==1){printf("%lld\n",k);continue;}
memset(pan,0,sizeof(pan));
long long f=phi(m);

// for(i=1;i<=m;i++)printf("%lld:%lld\n",i,pan[i]);
long long t=(k-1)/f;
long long p=(k-1)%f+1;
long long cnt=0;
for(i=1;i<=1000001;i++)
{
if(pan[i]==0)cnt++;
if(cnt==p){cnt=i;break;}
}
long long sum=cnt+t*m;
printf("%lld\n",sum);
}
return 0;
}