Sweet Problem CodeForces - 1263A

You have three piles of candies: red, green and blue candies:

the first pile contains only red candies and there are r candies in it,
the second pile contains only green candies and there are g candies in it,
the third pile contains only blue candies and there are b candies in it.
Each day Tanya eats exactly two candies of different colors. She is free to choose the colors of eaten candies: the only restriction that she can’t eat two candies of the same color in a day.

Find the maximal number of days Tanya can eat candies? Each day she needs to eat exactly two candies.

Input
The first line contains integer t (1≤t≤1000) — the number of test cases in the input. Then t test cases follow.

Each test case is given as a separate line of the input. It contains three integers r, g and b (1≤r,g,b≤108) — the number of red, green and blue candies, respectively.

Output
Print t integers: the i-th printed integer is the answer on the i-th test case in the input.

Example
Input
6
1 1 1
1 2 1
4 1 1
7 4 10
8 1 4
8 2 8
Output
1
2
2
10
5
9
Note
In the first example, Tanya can eat candies for one day only. She can eat any pair of candies this day because all of them have different colors.

In the second example, Tanya can eat candies for two days. For example, she can eat red and green candies on the first day, and green and blue candies on the second day.

In the third example, Tanya can eat candies for two days. For example, she can eat red and green candies on the first day, and red and blue candies on the second day. Note, that two red candies will remain uneaten.

题意：
有3种颜色的糖果若干种。你每天可以选两种不同颜色的糖果吃。请问最多可以吃多少天。
思路：
假设3种糖果数量a ≤ b ≤ c.
当c ≥ a + b时，每天一个c加一个a(b)最优，最多只能吃a + b天。
当c < a + b时，你可以设置一个最优的分配方法，前c天每天一个c加一个a(b)，后几天每天一个a,b，最多可以达到c + (a + b - c) / 2，至于怎样最优我们不关心，天数反正能算出来

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

int a[10];
int main()
{
    int T;scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d%d",&a[1],&a[2],&a[3]);
        sort(a + 1,a + 1 + 3);
        if(a[3] >= a[1] + a[2])
        {
            printf("%d\n",a[1] + a[2]);
        }
        else
        {
            printf("%d\n",a[3] + (a[1] + a[2] - a[3]) / 2);
        }
    }
    return 0;
}