# Codeforces Round #603 (Div. 2) A. Sweet Problem 水题

## A. Sweet Problem

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

给你三个数，每次你可以选择其中两个数-1，然后问你最多减多少次，使得所有数都大于等于0

## 题解

视频题解 https://www.bilibili.com/video/av77514280/

其实答案就是min(a+b,(a+b+c)/2)，考虑a+b和c的大小关系即可

## 代码

```
#include<bits/stdc++.h>
using namespace std;
long long a[3];
void solve(){
for(int i=0;i<3;i++)
cin>>a[i];
sort(a,a+3);
cout<<min(a[0]+a[1],(a[0]+a[1]+a[2])/2)<<endl;
}
int main(){
int t;
cin>>t;
while(t--)solve();
}
```