输入两个整数m和n，及另一个整数k，计算m/n，结果精确到小数点后k位。

#include<stdio.h>
int main()
{
int m,n,k,i;
printf("Please input integer m , n and k\n");
scanf("%d%d%d",&m,&n,&k);
printf("%d.",m/n);
for(i=1;i<k;i++)
{
m=m%n;
m*=10;
printf("%d",m/n);
}
m=m%n;
m*=10;
if(m%n<0.5*n)
printf("%d\n",m/n);
else
printf("%d\n",m/n+1);
getchar();
getchar();
return 0;
}

Private Sub Command1_Click()
Dim m As Integer, n As Integer, k As Integer, a As Single
m = InputBox("请输入整数m:")
Print "整数m的值:" & m
n = InputBox("请输入整数n:")
Print "整数n的值:" & n
k = InputBox("请输入正整数k:")
Print "正整数k的值:" & k
a = Round(m / n, k)
Print "m/n,k位小数的值:" & a
End Sub

#include <iostream>
#include <iomanip>
using namespace std;
int main(int argc, char* argv[]){
int m,n,p;
double k;
cin>>m>>n>>p;
k=(float)m/n;
cout.precision(p); // 精度	cout<<k<<endl;
return 0;
}