如题所述
double n,t,sum,pi;
sum=1.0;
n=1.0
t=1.0;
do{
t=t*(n/(2*n+1));
sum=sum+t;
}
while(n/(2*n+1)<0.0001);
pi=2*sum;
sum=1.0;
n=1.0
t=1.0;
do{
t=t*(n/(2*n+1));
sum=sum+t;
}
while(n/(2*n+1)<0.0001);
pi=2*sum;
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