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

URI Online Judge Solution URI Solution - Data Structure

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

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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