关于C++习题

求π/2近似值的公式为π/2=2/1*2/3*4/3*4/5*...*2n/(2n-1)*2n/(2n+1)

式中n=1,2,3...设计一个程序，求出n=1000时π的近似值

方法一：
double m, k = 1;
double n;
for (n = 1; n <= 1000; n++)
{
m = (2 * n / (2 * n - 1)) * (2 * n / (2 * n + 1));
k *= m;
m = 1;
}
Console.Write("{0}", k);
Console.Read();

方法二：
double n, i = 1, j = 1, k = 1;
for (n = 1; n <= 1000; n++)
{
i = i * (2 * n);
j = j * (2 * n - 1);
k = k * (2 * n + 1);
}
double a;
a = i / (j * k);
Console.Write("{0},{1},{2},{3}", a, i, j, k);
Console.Read();

#include<math.h> #include<stdio.h>

void main() { double fenzi=1,fenmu=1; double sum=0,pi; while(fenmu<=100) { sum=sum+fenzi/fenmu; fenzi=-1*fenzi; fenmu=fenmu+2; } pi=sum*4; printf("%f",pi); }

#include <iostream> #include <cmath> using namespace std; int main() { int n = 2; int x = 1; double pi = 1; for ( n ; n <= 1000; n += 2 ) { pi = pi * ( (n*n*1.0) / ( x * (x + 2) ) ); x += 2; } cout << "π的近似值是 = " << pi*2 << endl; cout << "π的实际值是 = " << M_PI << endl; return 0; }

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