# [POJ 2773] Happy 2006【gcd】

• 2018-01-31
• 0
• 0

## Problem:

 Time Limit: 3000MS Memory Limit: 65536K

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
```

Source

POJ Monthly--2006.03.26,static

## Code: O(Tmlogm) [3808K, 2407MS]

```#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;

int m, k, rp[1000005], rpcnt;

inline int gcd(int a, int b) {return !b ? a : gcd(b, a % b);}

int main(){
while(scanf("%d%d", &m, &k) != EOF){
rpcnt = 0, k--;
for(register int i = 1; i <= m; i++)
if(gcd(i, m) == 1) rp[rpcnt++] = i;
ll ans = rp[k % rpcnt] + m * (k / rpcnt);
printf("%lld\n", ans);
}
return 0;
}
```

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