【CodeForces】Round #345 (Div. 2) A. Joysticks（水）

题目链接：点击打开

A. Joysticks

题目连接：

http://www.codeforces.com/contest/651/problem/A

Description

Friends are going to play console. They have two joysticks and only one charger for them. Initially first joystick is charged at a1 percent and second one is charged at a2 percent. You can connect charger to a joystick only at the beginning of each minute. In one minute joystick either discharges by 2 percent (if not connected to a charger) or charges by 1 percent (if connected to a charger).

Game continues while both joysticks have a positive charge. Hence, if at the beginning of minute some joystick is charged by 1 percent, it has to be connected to a charger, otherwise the game stops. If some joystick completely discharges (its charge turns to 0), the game also stops.

Determine the maximum number of minutes that game can last. It is prohibited to pause the game, i. e. at each moment both joysticks should be enabled. It is allowed for joystick to be charged by more than 100 percent.

Input

The first line of the input contains two positive integers a1 and a2 (1 ≤ a1, a2 ≤ 100), the initial charge level of first and second joystick respectively.

Output

Output the only integer, the maximum number of minutes that the game can last. Game continues until some joystick is discharged.

Sample Input

3 5

Sample Output

6

这里要注意中间的有下划线和红字的一段话（翻译：如果有电量为1%的话，就要插充电器，否则就Gameover），如果a和b都等于1的情况就出现了矛盾，这时候就需要退出了。周赛就是错到了这里，好伤心T ^ T。

就是给电量少的充电就行了。

代码如下：

#include <cstdio>
#include <algorithm>
using namespace std;
int main()
{
	int a,b;
	int ans;
	while (~scanf ("%d %d",&a,&b))
	{
		ans = 0;
		while (a>0 && b>0)
		{
			if (a == 1 && b == 1)
				break;
			if (a<b)
			{
				a ++;
				b -= 2;
			}
			else
			{
				a -= 2;
				b++;
			}
			ans++;
		}
		printf ("%d\n",ans);
	}
	return 0;
}