code chef - Cool Guys题解-优快云博客

本文详细介绍了如何通过编程解决问题：给定一个整数N，随机选择两个整数A和B（1到N之间），计算A和B的最大公约数等于B的概率。通过求解最大公约数的算法和数学公式，本文提供了完整的解决方案和代码实现。

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All submissions for this problem are available.

Given an integerN. IntegersAandBare chosen randomly in the range[1..N]. Calculate the probability that the Greatest Common Divisor(GCD) ofAandBequals toB.

Input

The first line of the input contains an integerTdenoting the number of test cases. The description ofTtest cases follows. Each test case consists of a single integerNon a separate line.

Output

For each test case, output a single line containing probability as an irreducible fraction.

Given an integerN. IntegersAandBare chosen randomly in the range[1..N]. Calculate the probability that the Greatest Common Divisor(GCD) ofAandBequals toB.

Input

The first line of the input contains an integerTdenoting the number of test cases. The description ofTtest cases follows. Each test case consists of a single integerNon a separate line.

Output

For each test case, output a single line containing probability as an irreducible fraction.

Example

Input:
3
1
2
3

Output:
1/1
3/4
5/9

Constraints

1<=T<=10³

1<=N<=10⁹

本题也是个数学题。

有两个知识点：

1 求最大公约数的算法

2 求小于等于一个数值的两数相乘的对数 - 这个又是数学公式

据说这里是数学详细解析，感兴趣的研究一下：

http://matwbn.icm.edu.pl/ksiazki/mon/mon42/mon4204.pdf

OJ系统：

http://www.codechef.com/problems/COOLGUYS/

long long mGcd(long long a, long long b)
{
	long long c = 0;
	while (b)
	{
		c = b;
		b = a % b;
		a = c;
	}
	return c;
}

long long pairsOfCoolguys(long long n)
{
	long long ans = 0;
	long long sq = (long long)sqrt(n);
	for (unsigned i = 1; i <= sq; i++)
	{
		ans += n/i;
	}
	ans = (ans<<1) - sq*sq;
	return ans;
}

void coolguys()
{
	long long n = 0;
	int T = 0;
	cin>>T;
	while (T--)
	{
		cin>>n;
		long long ps = pairsOfCoolguys(n);
		n *= n;
		long long d = mGcd(ps, n);
		cout <<(ps/d) << "/"<<(n/d)<<"\n";
	}
}