1. 程式人生 > >輸入兩個整數m和n,及另一個整數k,計算m/n,結果精確到小數點後k位。

輸入兩個整數m和n,及另一個整數k,計算m/n,結果精確到小數點後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("請輸入整數m:")
Print "整數m的值:" & m
n = InputBox("請輸入整數n:")
Print "整數n的值:" & n
k = InputBox("請輸入正整數k:")
Print "正整數k的值:" & k
a = Round(m / n, k)
Print "m/n,k位小數的值:" & 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;
}

歡迎大神們前來補充啊~