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

题目描述

时间限制：1s；空间限制：256MB

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.

输入

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.

输出

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

样例

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

题意分析

q组数据，每组给定三个正数，每次从中选两个数减一，问最多经过几次使得这三个数中第一次出现0

C++代码

#include <cstdio>
#include <algorithm>

using namespace std;

int main() {
    int q;
    scanf("%d", &q);
    while (q --) {
        int r, g, b;
        scanf("%d%d%d", &r, &g, &b);
        // 最小值，最大值，中间值
        // 思路就是将中间值分配到最小值和最大值上，使得这两个值的大小尽可能接近 
        int mi = min(r, min(g, b)), ma = max(r, max(g, b)), mid = r + g + b - mi - ma;
        if (mi + mid <= ma) mi += mid;
        else {
            int tmp = mi + mid - ma;
            mi += mid;
            mi -= tmp / 2 + tmp % 2;
            ma += tmp / 2;
        }
        printf("%d\n", mi);
    }
    return 0;
}