Algorithm
Problem Name: beecrowd | 2310
Problem Link: https://www.beecrowd.com.br/judge/en/problems/view/2310
Volleyball
By Leonardo Fernandes, IFSC 
Brazil
Timelimit: 1
A volleyball coach would like to keep statistics about his team. In every game he keeps track of the number of services, blocks and attacks of each player, as well as how many of these services, blocks and attacks were successful (scored points). Your program must show, with two decimal points what is the total percentage of services, blocks and attacks by the whole team that were successful.
Input
Input starts with the number of players N (1 ≤ N ≤ 100), followed by the names of each of these players. Below the name of each player, two rows of integers are presented. In the first row, the numbers represent the service, blocks and attack attempts (0 ≤ S,B,A ≤ 10000) by the specific player. In the second row there is the number of these services, blocks and attacks (0 ≤ S1 ≤ S; 0 ≤ B1 ≤ B; 0 ≤ A1 ≤ A) that were successful.
Output
The output must contain the total percentage of successful services, blocks and attacks by the whole team, with two digits after the decimal point, as shown in the example.
| Input Sample | Output Sample |
|
3 |
Pontos de Saque: 19.05 %. |
Code Examples
#1 Code Example with C Programming
Code -
C Programming
#include <stdio.h>
struct info
{
char str[100];
int at[3];
int su[3];
};
int main(void)
{
int i, j, n, x=0, y=0, z=0, a=0, b=0, c=0;
float p, q, r;
scanf("%i", &n);
struct info volly[n];
for (i = 0; i < n; ++i)
{
scanf("%s", volly[i].str);
for (j = 0; j < 3; ++j)
scanf("%i", &volly[i].at[j]);
for (j = 0; j < 3; ++j)
scanf("%i", &volly[i].su[j]);
}
for (i = 0; i < n; ++i)
{
for (j = 0; j < 3; ++j)
{
if (j==0)
{
x+=volly[i].at[j];
a+=volly[i].su[j];
}
else if (j==1)
{
y+=volly[i].at[j];
b+=volly[i].su[j];
}
else if (j==2)
{
z+=volly[i].at[j];
c+=volly[i].su[j];
}
}
}
p=(a*100)/(float)x;
q=(b*100)/(float)y;
r=(c*100)/(float)z;
printf("Pontos de Saque: %.2f %%.\nPontos de Bloqueio: %.2f %%.\nPontos de Ataque: %.2f %%.\n", p, q, r);
return 0;
}
Input
Output
#2 Code Example with C++ Programming
Code -
C++ Programming
#include <iostream>
#include <string>
int main(){
int n, i;
double sbaatemp[4], sbasuc[4],totalsbaat[4] = {}, totalsbasu[4] = {};
std::string name, output[4] = {"Pontos de Saque: ", "Pontos de Bloqueio: ", "Pontos de Ataque: "};
std::cin >> n;
while (n--) {
std::cin >> name;
for (i = 0;i < 3;i++) {
std::cin >> sbaatemp[i];
totalsbaat[i] += sbaatemp[i];
}
for (i = 0;i < 3;i++) {
std::cin >> sbasuc[i];
totalsbasu[i] += sbasuc[i];
}
}
for (i = 0;i < 3;i++) {
std::cout << output[i];
std::cout.precision(2);
std::cout.setf(std::ios::fixed);
std::cout << (totalsbasu[i]*100)/totalsbaat[i] << " %." << std::endl;
//totalsbaat[i] = 100
//totalsbasu[i] = x
}
return 0;
}
Input
Output
#3 Code Example with Javascript Programming
Code -
Javascript Programming
var input = require('fs').readFileSync('/dev/stdin', 'utf8');
var lines = input.split('\n');
let number = parseInt(lines.shift());
let i = 0;
let SumA = 0;
let SumB = 0;
let SumS = 0;
let SumA1 = 0;
let SumB1 = 0;
let SumS1 = 0;
while(i < number){
let name = lines.shift();
let [S, B, A] = lines.shift().split(" ");
let [S1, B1, A1] = lines.shift().split(" ");
SumA += parseInt(A);
SumB += parseInt(B);
SumS += parseInt(S);
SumA1 += parseInt(A1);
SumB1 += parseInt(B1);
SumS1 += parseInt(S1);
i++;
}
console.log(`Pontos de Saque: ${((SumS1 * 100) / SumS).toFixed(2)} %.\nPontos de Bloqueio: ${((SumB1 * 100) / SumB).toFixed(2)} %.\nPontos de Ataque: ${((SumA1 * 100) / SumA).toFixed(2)} %.`)
Input
Output
#4 Code Example with Python Programming
Code -
Python Programming
n = int(input())
t = [0, 0, 0]
s = [0, 0, 0]
while n > 0:
n -= 1
c = input()
for i in range(1):
l = [int(x) for x in input().split()]
for x in range(3):
t[x] += l[x]
l = [int(x) for x in input().split()]
for x in range(3):
s[x] += l[x]
print('Pontos de Saque: %.2f %%.' % (100*(s[0]/t[0])))
print('Pontos de Bloqueio: %.2f %%.' % (100*(s[1]/t[1])))
print('Pontos de Ataque: %.2f %%.' % (100*(s[2]/t[2])))
Input
Output