Happy 2006
Time Limit:3000MS Memory Limit:65536KB 64bit IO Format:%I64d
& %I64u
Submit Status Practice POJ
2773
Appoint description:
System Crawler (2015-04-06)
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：给定两个数m,k，要求你求出第k个和m互质的数。
思路：因为对于公约数来说有gcd(a,b)=gcd(a+t*b,b)。所以在[1，m-1]中与m互素的个数与在[k*m+1，(k+1)*m-1]的互素的个数是一样的，即都等于phi(m),，所以只要找出[1，m-1]中与m互素的数，就能找到对用的接下来的区间中互素的数：那么答案就等于第k%phi(m)个与m互素的值p+m*(k/phi(m))。
#include <stdio.h>
#include <math.h>
#include <string.h>
#include <stdlib.h>
#include <iostream>
#include <sstream>
#include <algorithm>
#include <set>
#include <queue>
#include <stack>
#include <map>
using namespace std;
typedef long long LL;
const int inf=0x3f3f3f3f;
const double pi= acos(-1.0);
int prime[1000010];
int Euler(int n)//筛选素数+求<=n的与n互素的数
{
int i,j;
int m=n;//因为下面的n是要变得，所以先将n存起来
int c=n;//返回的是pji[m]
memset(prime,0,sizeof(prime));
for(i=2;i*i<=n;i++){
if(n%i==0){
n/=i;
for(j=1;i*j<=m;j++)//用筛选法标记所有与n不互素的数
prime[i*j]=1;
c=c/i*(i-1);
while(n%i==0)
n=n/i;
}
}
if(n>1){
for(j=1;n*j<=m;j++)// 当n>1的时候，n本身也是一个素因子
prime[n*j]=1;
c=c/n*(n-1);
}
return c;
}
int main()
{
int m,k,i;
int cnt;
int sum;
int t;
while(~scanf("%d %d",&m,&k)){
cnt=Euler(m);
t=k/cnt;
sum=0;
if(k%cnt==0)
t--;
k=k-cnt*t;
for(i=1;i<=m;i++){
if(!prime[i])
sum++;
if(sum==k)
break;
}
printf("%d\n",m*t+i);
}
return 0;
}