NC207581. SumoandCoins
描述
输入描述
The first line of the input is a single integer, T test cases follow.
Each test case has three integers, a, b(a+b=n).
输出描述
If all the coins can face up or down, output "ALL"(without quotes);
If all the coins can only face up, output "UP"(without quotes);
If all the coins can only face down, output "DOWN"(without quotes);
If all the coins cannot face the same side, output "NULL"(without quotes).
示例1
输入:
1 2 1 1
输出:
ALL
C++14(g++5.4) 解法, 执行用时: 36ms, 内存消耗: 608K, 提交时间: 2020-06-06 15:43:46
#include<bits/stdc++.h> using namespace std; #define ll long long int main(){ int t,n,a,b; cin>>t; while(t--){ cin>>n>>a>>b; if(n%2==0) cout<<"ALL"<<endl; else{ if(a%2==1) cout<<"UP"<<endl; else cout<<"DOWN"<<endl; } } }
C(clang 3.9) 解法, 执行用时: 6ms, 内存消耗: 376K, 提交时间: 2020-10-10 21:36:57
#include<stdio.h> int main() { int t,n,a,b; scanf("%d",&t); while(t--) { scanf("%d%d%d",&n,&a,&b); if(a % 2 == b % 2) printf("ALL\n"); else{ if(a % 2 == 1) printf("UP\n"); else printf("DOWN\n"); } } }
C++11(clang++ 3.9) 解法, 执行用时: 7ms, 内存消耗: 488K, 提交时间: 2020-06-06 16:39:15
#include<stdio.h> int main() { int t,n,a,b; scanf("%d",&t); while (t--) { scanf("%d%d%d",&n,&a,&b); if (n%2==0) printf("ALL\n"); else if (a&1) printf("UP\n"); else printf("DOWN\n"); } return 0; }
Python3(3.9) 解法, 执行用时: 77ms, 内存消耗: 2936K, 提交时间: 2021-03-23 16:18:50
for _ in range(int(input())): n,a,b=map(int,input().split()) if n%2==0: print('ALL') elif a%2==1: print('UP') else: print('DOWN')