URI Online Judge Solution 1023 Drought - Solution in C, C++, Java, Python and C#

URI Online Judge Solution 1023 Drought  - Solution in C, C++, Java, Python and C#

URI Online Judge Solution 1023 Drought   | Beginner
URI Problem Link - https://www.urionlinejudge.com.br/judge/en/problems/view/1023

Problem Name: 1023 Drought  solution
Problem Number : URI - 1023 Drought  code
Online Judge : URI Online Judge Solution
Category: Data Structure
Solution Language : C,C plus plus, java, python, c#(c sharp)

URI 1023 problem simple input output:

Input Sample	Output Sample
3 3 22 2 11 3 39 5 1 25 2 20 3 31 2 40 6 70 0	Cidade# 1: 2-5 3-7 3-13 Consumo medio: 9.00 m3. Cidade# 2: 5-10 6-11 2-20 1-25 Consumo medio: 13.28 m3.

URI Solution 1023 Drought   Code in C++ / URI 1023 code in CPP:

#include <cstdio>
#include <cstring>
#include <cmath>

using namespace std;

int arr[300];

int main(int argc, char const *argv[])
{
 int i, j, n, c = 1, a, b, ta, tp, fp;
 double ip;
 bool bo = false;

 while(scanf("%d", &n) && n)
 {
  if(bo) printf("\n");
  bo = true;

  ta = tp = 0;
  memset(arr, 0, sizeof arr);

  for (i = 0; i < n; ++i)
  {
   scanf("%d %d", &a, &b);
   ta += b;
   tp += a;
   arr[b/a] += a;
  }

  printf("Cidade# %d:\n", c); c++;

  for(i = 0, j = 0; i < 300; i++)
  {
            if(arr[i] > 0){
                if(j != 0)
                    printf(" ");
                printf("%d-%d", arr[i], i);
                j++;   
            }  
        }
  printf("\n");

  fp = (int) (modf ((double)ta/tp, &ip) * 100);

  if(fp < 10) printf("Consumo medio: %d.0%d m3.\n", (int)ip, (int)fp);
  else printf("Consumo medio: %d.%d m3.\n", (int)ip, (int)fp);
 }

 return 0;
}

URI Solution 1023 Drought   Code / URI 1002 solution in Java:

URI Solution 1023 Drought   Code / URI 1002 solution in  Python:

URI Solution 1023 Drought   Code / URI 1002 solution in  C# (C Sharp):

Demonstration:

Just implement this in coding. Since having any problem just put a comment below. Thanks



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