Solve of URI 1006 (Average 2) in C & C++
Problem Description of URI Online Judge 1006:
Problem name: Average 1.
Problem level: 1.
Problem category: Beginner.
Points: 1.4 points
Base time limit: 1 second.
Memory limit: 200 MB.
Problem subject: Sequential.
Problem adapted by: Neilor Tonin (Brazil).
Link: https://www.urionlinejudge.com.br/judge/en/problems/view/1005
Solution language: C & C++
My profile: https://www.urionlinejudge.com.br/judge/en/profile/38770
Solution of URI 1006 in C:
//URI online judge 1006 solution in C
//This solution is made and uploaded by Mehedi Hasan Kajol
#include<stdio.h>
int main(void){
float a, b, c, average;
scanf("%f%f%f", &a, &b, &c);
average = ( a*2.0 + b*3.0 + c*5.0) / 10.0;
printf("MEDIA = %.1f\n", average);
return 0;
}
#include<stdio.h>
int main(void){
float a, b, c, average;
scanf("%f%f%f", &a, &b, &c);
average = ( a*2.0 + b*3.0 + c*5.0) / 10.0;
printf("MEDIA = %.1f\n", average);
return 0;
}
Solution of URI 1006 in C++:
//This solution is made and uploaded by Mehedi Hasan Kajol
#include<iostream>
using namespace std;
int main(void){
float a, b, c, average;
cin >> a >> b >>c;
average = ( a*2.0 + b*3.0 + c*5.0 ) / 10.0;
printf("MEDIA = %.1f\n", average); //printf() for precision.
return 0;
}
#include<iostream>
using namespace std;
int main(void){
float a, b, c, average;
cin >> a >> b >>c;
average = ( a*2.0 + b*3.0 + c*5.0 ) / 10.0;
printf("MEDIA = %.1f\n", average); //printf() for precision.
return 0;
}