补题D题（Eat Candies）

最新推荐文章于 2022-04-28 17:43:10 发布

Jas.

最新推荐文章于 2022-04-28 17:43:10 发布

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本文链接：https://blog.youkuaiyun.com/qq_45793174/article/details/103841653

本文探讨了在有限数量的红、绿、蓝三种颜色糖果中，如何制定策略使每天能吃到两种不同颜色糖果的时间最大化。通过排序和差值计算，提出了两种解决策略：一是当任意两堆糖果的差值大于第三堆糖果数量时，直接消耗较小两堆糖果；二是当差值小于第三堆糖果数量时，先均衡消耗差值部分，再将剩余糖果平均分配，确保每天都能吃到不同颜色的糖果。

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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
输入三堆糖果数量，每天吃两块不相同的，求能每天吃两块不同糖果的最多天数。

思路
首先对三堆糖果进行排序，然后对a[1]和a[2]做差并和a[0]比较，如果差数量更多，将第一天糖果吃完，然后加上a[1]就是结果；如果第一堆糖果数量更多，将a[0]和a[2]分别减去差，然后将a[0]均分两份，分别和a[1]和a[2]一起吃，最后同时吃完。

#include <iostream>
#include <bits/stdc++.h>
using namespace std;
int main()
{
 int a[3];
 int x;
 int n,i;
  cin>>n;
 int count;
 while(n--){
  count = 0;
  memset(a,0,sizeof(a));
  cin>>a[0]>>a[1]>>a[2];
  sort(a,a+3);//升序 
  int k;
  k = a[2] -a[1];
  if(k >= a[0]){
 //第一种情况呢 1 6 100000 或 1 6 7 
  count+=a[0]+a[1];
  }
 //第二种情况  3 4 5  /  2 8 8
  else{
   count+=k;
   a[2]-=k;
   a[0]-=k;
   if(a[0]>1){
    x=a[0]/2;
    a[1]-=x;
    a[2]-=x;
    count+=2*x;}
  count+=a[1];   
  } 
  cout<<count<<endl;
 }
}