A. Shovels and Swords

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这篇博客探讨了一款计算机游戏中的资源管理问题。玩家Polycarp可以通过合成铲子和剑来获取绿宝石。每把剑需要2个钻石和1根木棍，每个铲子需要2个木棍和1个钻石。玩家的目标是通过最优的合成策略来最大化绿宝石的数量。文章提供了一个简单的算法，根据木棍和钻石的相对数量来确定最佳合成方案，并给出了若干示例测试用例来说明这个策略的运用。

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A. Shovels and Swords
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
Polycarp plays a well-known computer game (we won’t mention its name). In this game, he can craft tools of two types — shovels and swords. To craft a shovel, Polycarp spends two sticks and one diamond; to craft a sword, Polycarp spends two diamonds and one stick.

Each tool can be sold for exactly one emerald. How many emeralds can Polycarp earn, if he has a sticks and b diamonds?

Input
The first line contains one integer t (1≤t≤1000) — the number of test cases.

The only line of each test case contains two integers a and b (0≤a,b≤109) — the number of sticks and the number of diamonds, respectively.

Output
For each test case print one integer — the maximum number of emeralds Polycarp can earn.

Example
inputCopy
4
4 4
1000000000 0
7 15
8 7
outputCopy
2
0
7
5
Note
In the first test case Polycarp can earn two emeralds as follows: craft one sword and one shovel.

In the second test case Polycarp does not have any diamonds, so he cannot craft anything.

#include<iostream>
using namespace std;
int main()
{
	int n,a,b;
	cin>>n;
	while(n--)
	{
		    cin>>a>>b;
		    if(a<(a+b)/3)
		    cout<<a<<endl;
		    else
		    if(b<(a+b)/3)
		    cout<<b<<endl;
		    else
		    cout<<(a+b)/3<<endl;
	}
}