URI Problem 1005 Average 1 Solution in C

Code:  

#include<stdio.h>
int main()
{
    double a,b,media;
    scanf("%lf %lf",&a,&b);
    media=(((a*3.5)+(b*7.5))/(3.5+7.5));
    printf("MEDIA = %.5lf\n",media);
    return 0;
}     

 

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URI online judge problem 1004

Code: 

#include<stdio.h>
int main()
{
    int a,b,product;
    scanf("%d %d",&a,&b);
    product=a*b;
    printf("PROD = %d\n",product);
    return 0;
}

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URI problem 1003

Code:

#include<stdio.h>
int main()
{
    int A,B,sum;
    scanf("%d %d",&A,&B);
    sum=A+B;
    printf("SOMA = %d\n",sum);
    return 0; 
}

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URI problem 1002 Solution

Code:

#include <stdio.h>
int main()
{
    double A,R;
    scanf("%lf",&R);
    A = 3.14159 *(R * R);
    printf("A=%0.4lf\n",A);
    return 0;
}                    

                         

1180 Lowest Number and Position

                                                  URI Online Judge | 1180

              Lowest Number and Position

                                          Adapted by Neilor Tonin, URI Brazil

Timelimit: 1

Write a program that reads a number N. This N is the size of a array X[N]. Next, read each of the numbers of X, find the smallest element of this array and its position within the array, printing this information.

Input

The first line of input contains one integer N (1 < N <1000), indicating the number of elements that should be read to an array X[N] of integer numbers. The second row contains each of the N values, separated by a space.

Output

The first line displays the message “Menor valor:” followed by a space and the lowest number read in the input. The second line displays the message “Posicao:” followed by a space and the array position in which the lowest number is, remembering that the array starts at the zero position.

Input Sample 10 1 2 3 4 -5 6 7 8 9 10	Output Sample                                               Menor valor: -5                                     Posicao: 4
Solution: #include<stdio.h> int main() {     int N[1000],i,count=0,X;     int min=N[count];     scanf("%d",&X);     for(i=0;i<X;i++)     {         scanf("%d",&N[i]);     }     for(i=0;i<X;i++)     {         if(N[i]<min)         {             min=N[i];             count=i;         }     }     printf("Menor valor: %d\nPosicao: %d\n",N[count],count);       return 0; }