Noi OJ 1.3 05: floating point numeric C language for calculating fractions

describe

Two integers a and b As numerator and denominator respectively , Existing score a/b , Find its floating-point value （ Double precision floating point , After decimal point 9 position ）

Input

Enter only one line , Includes two integers a and b（b Not for 0）

Output

The output is only one line , fraction a/b The floating-point value of （ Double precision floating point , After decimal point 9 position ）

There are two solutions to the problem

The first is to set the input number to double The type of , So the number after rounding out will also be double Type of （ as follows ）

#include<stdio.h>
int main(){
double a,b;
scanf("%lf %lf",&a,&b);
printf("%.9lf",a/b);
return 0;
}

The second can be seen as a cast

#include<stdio.h>
int main(){
int a,b;
scanf("%d %d",&a,&b);
double c=a*1.0/b;
printf("%.9lf",c);
return 0;
}

notes ： there   a  must   *1.0, If direct a/b Words , Both are int The integer of , So the number that comes out of the division must be an integer , But if we   *1.0, So we give integers a new type , You can go to double Divide by （ At this time a*1.0 Can be seen as a new double Number of types ）.