c语言中如何判断计算结果精确到小数点后8位 c语言怎样精确到小数点后8位.
C\u8bed\u8a00\u7f16\u7a0b\u8ba1\u7b97\u81ea\u7136\u6570\u5e73\u65b9\u7684\u5012\u6570\u7ec4\u6210\u7684\u7ea7\u6570\uff0c\u7cbe\u786e\u5230\u5c0f\u6570\u70b9\u540e8\u4f4d\uff1f#include
int main()
{
int i;
double sum;
for(i=1,sum=0;1.0/(i*i)>=1e-8;++i)
sum+=1.0/(i*i);
printf("%.8lf\n",sum);
return 0;
}
s \u548ci\u662fint\u7c7b\u578b\u7684\uff0c3/2 \u662f=1\u7684\uff0c\u5982\u679c\u662ffloat \u6216\u8005double 3.0/2.0=1.5\u7684
int multiply(int i){
if (i ==1 || i == 0) return 1;
else return (i*multiply(i-1));
}
int _tmain(int argc, _TCHAR* argv[])
{
int i=1,j=1;
double x;
scanf("%lf",&x);
double result=1;
while (1)
{
j = j*(-1);
int sum = multiply(i);
double k;
k=pow(x,2*i)/sum;
result =result+j*pow(x,2*i)/(double)sum;
if (k<=0.00000001)
{printf("result = %.8f ; i = %d",result, i);
break;}
i++;
}
system("pause");
return 0;
}
太笼统了 不好说 结果用%.8f就行了吧 注意%后面有个小数点
输出时，用％．8f即可
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