CodeForces 651 A Joysticks-优快云博客

探讨两个充电状态不同的游戏控制器如何通过合理充电策略延长游戏时间，直到其中一个控制器完全耗尽电量。输入包括两个控制器的初始电量百分比，输出为游戏能持续的最大分钟数。

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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



#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
#include <math.h>
#include <stdio.h>

using namespace std;
long long int dp[205][205];
int n,m;
long long int dfs(int x,int y)
{
    //cout<<x<<" "<<y<<endl;
    if(dp[x][y]!=-1)
        return dp[x][y];
    if(x-2<0&&y-2>=0)
       dp[x][y]=dfs(x+1,y-2)+1;
    else if(x-2>=0&&y-2<0)
       dp[x][y]=dfs(x-2,y+1)+1;
    else
    {
       dp[x][y]=max(dfs(x+1,y-2),dfs(x-2,y+1));
       dp[x][y]++;
    }

    return dp[x][y];
}
int main()
{
    memset(dp,-1,sizeof(dp));
    for(int i=1;i<=100;i++)
        dp[i][0]=0,dp[0][i]=0;
    dp[1][1]=0;
    scanf("%d%d",&n,&m);
    printf("%lld\n",dfs(n,m));
    return 0;
}