CodeForces 651A Joysticks 贪心

最新推荐文章于 2019-08-18 18:05:00 发布

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最新推荐文章于 2019-08-18 18:05:00 发布

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原文链接：http://www.cnblogs.com/Noevon/p/5664800.html

本文介绍了一道关于两个游戏手柄交替充电的问题，通过贪心算法实现最优充电策略，以延长游戏时间。提供了完整的C语言代码示例。

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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

题目大意：有两个手柄，我们会给出他剩余的电量，每次充电一分钟会多百分之一的电量，不充点会掉百分之二的电。
思路：贪心，优先选择电量少的充，注意两个手柄电量为1时特殊情况

#include <stdio.h>
int main()
{
    int a ,b, t=0;
    scanf("%d%d", &a, &b);
    while(a>=1&&b>=1)
    {
        if(a==1&&b==1)
        break;
        if(a>b)
            {
            a -= 2;
            b++;
            t++;
            }
        else
            {
            b -= 2;
            a++;
            t++;
            }
    }
    printf ("%d\n", t);
    return 0;
}