sicily Computing PI

DescriptionTime Limit: 1sec Memory Limit:256MB
You can approximate PI by using the following series:
PI = 4(1 - 1/3 + 1/5 - 1/7 + 1/9 - ... -1/(2i-1) + 1/(2i+1))

Input

An integer i (0<i<=100000).

Output

The PI value.

You should set the precision to 15.

Sample Input

Copy sample input to clipboard

10

Sample Output

3.232315809405593

一開始老是超時，最後放棄了按順序疊加的方法（因爲那要頻繁計算一個pow(-1,n)），而是把加法看成兩段，奇偶分開算，就沒超時了。

#include 
#include 

using namespace std;

double danshu(int lev)
{
	double sum=0;
	for (int i = 1; i <= lev; ++i)
	{
		sum += 1.0/(4*i - 3);
	}
	return sum;
}

double oushu(int lev)
{
	double sum=0;
	for (int i = 1; i <= lev; ++i)
	{
		sum += 1.0/(4*i - 1);
	}
	return -sum;
}

int main()
{
	int n,dan,shuang;
	cin>>n;
	double PI=0;
	n++;
	if (n % 2 ==0)
	{
		dan = n/2;
		shuang = dan;
	}
	else
	{
		shuang = n / 2;
		dan = shuang + 1;
	}
	cout<