Problem - A - Codeforces
codeforces.com
- 제목
Two Elevators
- 조건
time limit per test : 1 second
memory limit per test : 256 megabytes
input : standard input
output : standard output
- 문제
Vlad went into his appartment house entrance, now he is on the 1-th floor. He was going to call the elevator to go up to his apartment.
There are only two elevators in his house. Vlad knows for sure that:
the first elevator is currently on the floor a (it is currently motionless),
the second elevator is located on floor b and goes to floor c (b≠c). Please note, if b=1, then the elevator is already leaving the floor 1 and Vlad does not have time to enter it.
If you call the first elevator, it will immediately start to go to the floor 1. If you call the second one, then first it will reach the floor c and only then it will go to the floor 1. It takes |x−y| seconds for each elevator to move from floor x to floor y.
Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator.
- 입력
The first line of the input contains the only t (1≤t≤104) — the number of test cases.
This is followed by t lines, three integers each a, b and c (1≤a,b,c≤108, b≠c) — floor numbers described in the statement.
- 출력
Output t numbers, each of which is the answer to the corresponding test case.
As an answer, output:
- if it is better to call the first elevator;
- if it is better to call the second one;
- if it doesn't matter which elevator to call (both elevators will arrive in the same time).
| 예제 입력1 | 예제 출력1 |
| 3 1 2 3 3 1 2 3 2 1 |
1 3 2 |
#include <iostream>
#include <algorithm>
#include <cmath>
using namespace std;
#define endl '\n'
#define fastio ios::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
int main() {
fastio;
int sz, a, b, c;
cin >> sz;
while(sz--){
cin >> a >> b >> c;
int t = abs(b - c) + c - 1;
if(a - 1 < t) cout << 1 << endl;
else if (a - 1 > t) cout << 2 << endl;
else cout << 3 << endl;
}
return 0;
}