求1/1+1/3+1/5+...+1/99的和 , 求1/2+1/4+1/6+...+1/00的和

#include <iostream.h>

void main()
{
int n,m;
double sum1, sum2;

sum1 = 0; // 求1/1+1/3+1/5+...+1/99的和
for (n=0;n<=49;n++)
sum1 += 1.0/(2*n+1);
cout<<sum1<<endl;
sum2 = 1;// 求1/2+1/4+1/6+...+1/00的和
for(m=1;m<=50;m++)
sum2 +=1.0/(2*m);
cout<<sum2<<endl;
}