August Long Challenge Olympics Ranking Solution

Olympics Ranking Solution

In Olympics, the countries are ranked by the total number of medals won. You are given six integers G1G1, S1S1, B1B1, and G2G2, S2S2, B2B2, the number of gold, silver and bronze medals won by two different counties respectively. Determine which country is ranked better on the leaderboard. It is guaranteed that there will not be a tie between the two countries.

Also See: August Long Challenge 2021 Solutions

Input Format

• The first line of the input contains a single integer TT denoting the number of test cases. The description of TT test cases follows.
• The first and only line of each test case contains six space-separated integers G1G1, S1S1, B1B1, and G2G2, S2S2, B2B2.

Output Format

For each test case, print "1" if the first country is ranked better or "2" otherwise. Output the answer without quotes.

Constraints

• 1≤T≤10001≤T≤1000
• 0≤G1,S1,B1,G2,S2,B2≤300≤G1,S1,B1,G2,S2,B2≤30

Subtasks

Subtask #1 (100 points): Original constraints

Sample Input 1 

3
10 20 30 0 29 30
0 0 0 0 0 1
1 1 1 0 0 0

Sample Output 1 

1
2
1

Explanation

Test case 11: Total medals for the first country are 10+20+30=6010+20+30=60 and that for the second country are 0+29+30=590+29+30=59. So the first country is ranked better than the second country.

Test case 22: Total medals for the first country are 0+0+0=00+0+0=0 and that for the second country are 0+0+1=10+0+1=1. So the second country is ranked better than the first country.

Solution

Program Python:

#cook your dish here
t=int(input())
while (t > 0):
    g1,s1,b1,g2,s2,b2=map(int, input().split())
    sum1=g1+s1+b1
    sum2=g2+s2+b2
    if (sum1 > sum2):
        print(1)
    else:
        print(2)
    t=t-1

#include<bits/stdc++.h>
using namespace std;
long long int mod = 1e9+7;

int main()
{
  int t;
  cin>>t;

  while(t--)
  {
    int a=0,b=0;
    for(int i=0;i<3;i++)
    {
      int x;
      cin>>x;
      a+=x;
    }
    for(int i=0;i<3;i++)
    {
      int x;
      cin>>x;
      b+=x;
    }
    a>b?cout<<"1"<<endl : cout<<"2"<<endl;
  }
  return 0;
}