链接：

https://codeforces.com/contest/1263/problem/A

题意：

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.

思路：

排序得到r >= g >= b
考虑r >= g+b，答案是g+b
否则让r，g，b三个相等
首先让r=g，r减去r-g， b减去r-g（保证可行，r < g+b => b > r-g)
再让r = g = b,r和g同时减去g-(b-(r-g)),三者最后为b+g-r
考虑三个相等，即可。

代码：

#include<bits/stdc++.h>
using namespace std;

int main()
{
    int a[3];
    int t;
    cin >> t;
    while(t--)
    {
        for (int i = 0;i < 3;++i)
            cin >> a[i];
        sort(a, a+3);
        if (a[2] >= a[1]+a[0])
            cout << a[1]+a[0] << endl;
        else
        {
            int ans = 0;
            ans += (a[2]-a[1])+(a[2]-a[0]);
            int t = a[0]-a[2]+a[1];
            ans += (3*t)/2;
            cout << ans << endl;
        }
    }

    return 0;
}

原文地址：https://www.cnblogs.com/YDDDD/p/12053774.html