codeforces 651A Joysticks-优快云博客

本文介绍了一个关于双摇杆控制器的游戏耐久挑战问题，玩家需在只有一个充电器的情况下，通过合理分配充电来延长游戏时间。文章提供了一段C++代码实现，模拟了这一过程并计算出了游戏能够持续的最大分钟数。

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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.

   Examples
 
     input
   
3 5

     output
   
6

     input
   
4 4

     output
   
5

Note

In the first sample game lasts for 6 minute by using the following algorithm:

at the beginning of the first minute connect first joystick to the charger, by the end of this minute first joystick is at 4%, second is at 3%;
continue the game without changing charger, by the end of the second minute the first joystick is at 5%, second is at 1%;
at the beginning of the third minute connect second joystick to the charger, after this minute the first joystick is at 3%, the second one is at 2%;
continue the game without changing charger, by the end of the fourth minute first joystick is at 1%, second one is at 3%;
at the beginning of the fifth minute connect first joystick to the charger, after this minute the first joystick is at 2%, the second one is at 1%;
at the beginning of the sixth minute connect second joystick to the charger, after this minute the first joystick is at 0%, the second one is at 2%.

After that the first joystick is completely discharged and the game is stopped.

模拟这个过程，让电量少的+1，另一个-2，电量不足时就结束了！注意1和1，输出0

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