POJ 2773 Happy 2006 解题报告（容斥原理+质因数分解）

Happy 2006

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 9002		Accepted: 3013

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

    解题报告：素数的这些东西，做多了都像类型题了。不过这题考察了二分的使用，还是不错的题。另外注意1的情况。
    代码如下：

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

int prime[30];
int prime_num;

typedef long long LL;

LL co_prime(LL n)
{
LL ans = 0;

for(int i=1;i<(1<<prime_num);i++)
{
LL tmp = 1;
bool flag = false;

for(int j=0;j<prime_num;j++) if(i&(1<<j))
tmp = tmp*prime[j],flag = !flag;

ans += n/tmp*(flag?1:-1);
}

return n - ans;
}

int main()
{
int n;
LL k;

while(~scanf("%d%lld", &n, &k))
{
if(n==1)
{
printf("%lld\n", k);
continue;
}

prime_num = 0;
for(int i=2;i*i<=n;i++) if(n%i==0)
{
prime[prime_num++] = i;
while(n%i==0) n/=i;
}
if(n>1) prime[prime_num++] = n;

LL l = 1, r = 1ll<<40;
while(l<=r)
{
LL m = (l+r)/2;
if(co_prime(m)>=k)
r = m-1;
else
l = m+1;
}

printf("%lld\n", l);
}
}