C语言除法问题，关于保留小数的 怎样在c语言除法中保留小数点？
C\u8bed\u8a00\u9898\u76ee \u9664\u6cd5\u4fdd\u7559\u4e24\u4f4d\u5c0f\u6570#include
void main()
{
int i=15,j=4;
float h;
h=(float) (15*100/4)/100;
printf("%.2f",h);
}
\u6ce8\uff1a%f\uff1a\u4e0d\u6307\u5b9a\u5bbd\u5ea6\uff0c\u6574\u6570\u90e8\u5206\u5168\u90e8\u8f93\u51fa\u5e76\u8f93\u51fa6\u4f4d\u5c0f\u6570\u3002 %m.nf\uff1a\u8f93\u51fa\u5171\u5360m\u5217\uff0c\u5176\u4e2d\u6709n\u4f4d\u5c0f\u6570\uff0c\u5982\u6570\u503c\u5bbd\u5ea6\u5c0f\u4e8em\u5de6\u7aef\u8865\u7a7a\u683c\u3002
%-m.nf\uff1a\u8f93\u51fa\u5171\u5360n\u5217\uff0c\u5176\u4e2d\u6709n\u4f4d\u5c0f\u6570\uff0c\u5982\u6570\u503c\u5bbd\u5ea6\u5c0f\u4e8em\u53f3\u7aef\u8865\u7a7a\u683c\u3002
#include
int main()
{
float c;//\u9996\u5148\u8981\u628a\u4f60\u8981\u6c42\u7684\u6570\u5b9a\u4e49\u6210\u6d6e\u70b9\u578b\uff0c\u901a\u4fd7\u7684\u8bb2\u5c31\u662ffloat\u548cdouble\u5c31\u662f\u5c0f\u6570\u578b\uff1b
int a=5,b=3;//int\u6574\u6570\u578b\uff0c\u5b83\u81ea\u52a8\u628a\u5c0f\u6570\u53bb\u6389\uff1b
c=1.0*a/b;//\u6700\u91cd\u8981\u7684\u662f\u8981\u4e58\u4ee51.0\uff1b
printf("%.2f",c);//%.2f;%\u540e\u9762\u7684\u662f\u70b9\uff08.\uff09\u51e0\uff0c\u5c31\u4fdd\u7559\u51e0\u4f4d\u5c0f\u6570\uff1b
return 0;
}
\u5e0c\u671b\u80fd\u5e2e\u5230\u4f60\uff0c\u671b\u91c7\u7eb3\uff1b
保留小数点后100位。
#include <stdio.h>
#define N 100
int main()
{
int a[N],n=2,m=3,k,sum=0;
int i=0;
printf("请输入被除数n:");
scanf("%d",&n);
printf("请输入除数m:");
scanf("%d",&m);
k=n%m;
sum=n/m;
while(i<N)
{
k*=10;
a[i]=k/m;
k=k%m;
i++;
}
if(a[99]>=5) a[98]+=1;
printf("%d/%d=%d.",n,m,sum);
for(i=0;i<99;i++)
printf("%d",a[i]);
printf("
");
}
printf("%.10lf",a);
printf("%.7lf",a);
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