Spoj-BIPCSMR16 Team Building

To make competitive programmers of BUBT, authority decide to take regular programming contest. To make this contest more competitive and fruitful there are some rules given to balance a team:
1. Only 1st , 2nd and 3rd year student can participate. 
2. A team must have three members. 
3. All the member cannot be from same year. 
You need to find out the maximum number of teams can build up according to given rules.

Input

The first line of input contain an integer T (1<=T<=10000) test case. Next T line contains three positive integer X, Y and Z (1<=X, Y, Z<=2*10^9) separated by a space which denotes the number of participants from 1st, 2nd, and 3rd year student.

Output

You need to find out the maximum number of teams can build up according to given rules.

Example

Input
2
1 2 3
1 12 3

Output
2
4

不能取三个同年级的，就跟这题一模一样 cf478C

如果最大的人数比另外两种的人数之和小，就可以直接组成sum/3个

否则全部都是2:1的方案

 #include<cstdio>
 #include<iostream>
 #include<cstring>
 #include<cstdlib>
 #include<algorithm>
 #include<cmath>
 #include<queue>
 #include<deque>
 #include<set>
 #include<map>
 #include<ctime>
 #define LL long long
 #define inf 0x7ffffff
 #define pa pair<int,int>
 #define mkp(a,b) make_pair(a,b)
 #define pi 3.1415926535897932384626433832795028841971
 #define mod 1000000007
 using namespace std;
 inline LL read()
 {
     LL x=,f=;char ch=getchar();
     while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
     while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
     return x*f;
 }
 int main()
 {
     int T=read();
     while (T--)
     {
         LL a=read(),b=read(),c=read();
         printf("%lld\n",min((a+b+c)/,min(a+b,min(b+c,c+a))));
     }
 }

Spoj BIPCSMR16