CodeForces651AJoysticks

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

Input

3 5

Output

6

Input

4 4

Output

5

代码：

#include<stdio.h>#include<iostream>#include<string.h>#include<algorithm>#include<math.h>int min(int x,int y){return x<y?x:y;}int max(int x,int y){return x>y?x:y;}using namespace std;int main(){int a,b;while(scanf("%d %d",&a,&b)!=EOF){int sum=0,k,j;while(a>=1&&b>=1){if(a==1&&b==1){printf("%d\n",sum);break;}k=min(a,b);j=max(a,b);a=k;b=j;a++;b=b-2;sum++;} if(a==0||b==0)printf("%d\n",sum); } return  0;}

题意：有两个控制杆，你可以在每分钟最开始的时候控制充电器连接某个控制杆。每一分钟，连接控制杆的会冲入%1的电，没有连接的会失去%2的电。控制杆电量剩余%1的时候必须连接充电器。给你两个控制杆最开始的电量，问最多能坚持多少分钟。

思路：先判断出两个数的大小，给小的充电，大的失电。一直循环到最后某一方为0时或者两方都为1时结束（因为1时必须充电，当两方都为1时只有一个充电器，无法同时充电）。