【Codeforces Round 273 (Div 2)C】【贪心 脑洞】 Table Decorations 三种气球装饰桌子 同桌三个气球不完全同色的最多桌子装饰数

C. Table Decorations

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

You have r red, g green
and b blue balloons. To decorate a single table for the banquet you need exactly
three balloons. Three balloons attached to some table shouldn't have the same color. What maximum number t of
tables can be decorated if we know number of balloons of each color?
Your task is to write a program that for given values r, g and b will
find the maximum number t of tables, that can be decorated in the required manner.

Input
The single line contains three integers r, g and b (0 ≤ r, g, b ≤ 2·109)
— the number of red, green and blue baloons respectively. The numbers are separated by exactly one space.

Output
Print a single integer t —
the maximum number of tables that can be decorated in the required manner.

Sample test(s)

input

5 4 3

output

4

input

1 1 1

output

1

input

2 3 3

output

2

Note
In the first sample you can decorate the tables with the following balloon sets: "rgg",
"gbb", "brr", "rrg",
where "r", "g" and "b"
represent the red, green and blue balls, respectively.

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<string>
#include<ctype.h>
#include<math.h>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<bitset>
#include<algorithm>
#include<time.h>
using namespace std;
void fre(){freopen("c://test//input.in","r",stdin);freopen("c://test//output.out","w",stdout);}
#define MS(x,y) memset(x,y,sizeof(x))
#define MC(x,y) memcpy(x,y,sizeof(x))
#define MP(x,y) make_pair(x,y)
#define ls o<<1#define rs o<<1|1typedef long long LL;
typedef unsigned long long UL;
typedef unsigned int UI;
template <class T1,class T2>inline void gmax(T1 &a,T2 b){if(b>a)a=b;}
template <class T1,class T2>inline void gmin(T1 &a,T2 b){if(b<a)a=b;}
const int N=0,M=0,Z=1e9+7,ms63=1061109567;
LL a[5];
int main()
{
for(int i=1;i<=3;++i)scanf("%lld",&a[i]);
sort(a+1,a+4);
gmin(a[3],(a[1]+a[2])*2);
LL ans=(a[1]+a[2]+a[3])/3;
printf("%lld\n",ans);
return 0;
}
/*
【trick&&吐槽】
1，这种水题可是把zjs整整卡了一个一个小时，哇咔咔。
2，排序出来的结果，我竟然把a[1]当成了最大的，结果wa一发，好蠢！
还好最后还是3分钟就AC了。

【题意】
给你三种颜色的气球，数量分别为a[1],a[2],a[3]，数量都在[0,2e9]范围
我们想要用这三种颜色的气球装饰桌子。
每个桌子必须用三个气球来装饰，且三个气球的颜色不能完全相同。
问你，我们最多可以装饰多少个桌子。

【类型】
贪心 脑洞

【分析】
其实我们用自己博大的胸怀（噗）去想一想，就能想到。
不完全相同，这个限制条件，其实是很宽泛的。
如果三种颜色的气球，数量是比较接近的话，答案显然是sum/3
如果不接近，什么情况答案不是sum/3呢？

显然，是有一种气球的数量特别多的时候。
于是，我们排个序，使得a[1]<a[2]<a[3]，然后a[3]=min(a[3],a[1]+a[2]<<1)

接着输出sum/3就好啦~

*/