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
Output
Output the K-th element in a single line.
Sample Input
2006 1
2006 2
2006 3
Sample Output
1
3
5
Solution
题目大意:给定m和k求 与数m互质的第k个数。
解题思路:gcd(a,b)=gcd(a+k*b,b)
Code
#include
#include
using namespace std;
#define N 1000000
int prime[N];
int gcd(int a, int b)
{
return b ? gcd(b, a%b) : a;
}
int main()
{
int m, k;
while (~scanf("%d%d", &m, &k))
{
int res = 0;
for (int i = 1; i if (gcd(m, i) == 1)
prime[res++] = i;
if (k%res)
cout 1] else
cout 1)*m + prime[res - 1] return 0;
}
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转载文章请注明:POJ 2773 – Happy 2006(素数判定 欧几里得算法) - https://blog.i-ll.cc/archives/49