CF-478C Table Decorations (贪心)

Table Decorations

Time limit per test: 1 second

Memory limit per test: 256 megabytes

Problem Description

You have r red, g green and b blue balloons. To decorate a single table for the banquet you need exactly three balloons. Three balloons attached to some table shouldn't have the same color. What maximum number t of tables can be decorated if we know number of balloons of each color?

Your task is to write a program that for given values r, g and b will find the maximum number t of tables, that can be decorated in the required manner.

Input

The single line contains three integers r, g and b (0 ≤ r, g, b ≤ 2·10^9) — the number of red, green and blue balloons respectively. The numbers are separated by exactly one space.

Output

Print a single integer t — the maximum number of tables that can be decorated in the required manner.

Sample Input & output

5 4 3 (output: 4)

1 1 1 (output: 1)

2 3 3 (output: 2)

Note

In the first sample you can decorate the tables with the following balloon sets: "rgg", "gbb", "brr", "rrg", where "r", "g" and "b" represent the red, green and blue balls, respectively.

Solution

题目数据给的很大，显然不是模拟的意思。所以应是利用贪心思想。

思路是先对所给三个数值按降序排序，得到a、b、c。然后判断a与(b + c) * 2的关系：

若a >= (b + c) * 2，则拿两个a分别给b或c，则限制因素为b+c。所以最后解为b+c（a剩余多了）。如果是三种颜色都挑，则提高了题目的要求，也就是只需三个颜色不都相同就可以。

若a < (b+c) * 2，则限制因素为a。这时解为(a + b + c) / 3。解释是：首先可以发现在满足这个条件时，a是不够的，也就是说在a两个两个取完之前，b和c一定至少有一个还有剩。那么假设一个数据 3 3 2。模拟：1 2 2；0 0 2。也就是后面一定会剩下。那么是否会出现最后是0 1 6这种情况呢？也就是b的个数又成为限制因素。

我反推了试试：0 1 6；2 2 6；4 3 6；6 4 6；8 5 6；10 6 6（5个桌子）。

而实际上对于这个数据正确解法应是10 6 6；8 5 6；6 5 5；4 4 5；2 4 4；0 3 4；0 3 2；0 1 1（7个桌子）。

所以，0 1 6这种情况的出现是因为在中间的选择过程中就已经不满足贪心。

也就是说最后剩下的两个一定会是用到0 1、1 0、1 1三者之一，即剩下1个或2个。

所以整数除法(a + b + c) / 3 为正解。

if(a >= (b + c) * )
    cout << b + c << endl;
else
    cout << (a + b + c) /  << endl;