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。
阿新 • • 發佈:2019-01-03
#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~