标签:iostream http cstring pre nsis and das row ==
You may have already known that a standard ICPC team consists of exactly three members. The perfect team however has more restrictions. A student can have some specialization: coder or mathematician. She/he can have no specialization, but can‘t have both at the same time.
So the team is considered perfect if it includes at least one coder, at least one mathematician and it consists of exactly three members.
You are a coach at a very large university and you know that cc of your students are coders, mm are mathematicians and xx have no specialization.
What is the maximum number of full perfect teams you can distribute them into?
Note that some students can be left without a team and each student can be a part of no more than one team.
You are also asked to answer qq independent queries.
Input
The first line contains a single integer qq (1≤q≤1041≤q≤104) — the number of queries.
Each of the next qq lines contains three integers cc, mm and xx (0≤c,m,x≤1080≤c,m,x≤108) — the number of coders, mathematicians and students without any specialization in the university, respectively.
Note that the no student is both coder and mathematician at the same time.
Output
Print qq integers — the ii-th of them should be the answer to the ii query in the order they are given in the input. The answer is the maximum number of full perfect teams you can distribute your students into.
Example
6 1 1 1 3 6 0 0 0 0 0 1 1 10 1 10 4 4 1
1 3 0 0 1 3
Note
In the first example here are how teams are formed:
#include <iostream> #include <algorithm> #include <cstdio> #include <string> #include <cstring> #include <cstdlib> #include <map> #include <vector> #include <set> #include <queue> #include <stack> #include <cmath> using namespace std; #define ok return 0; typedef long long lli; #define TLE std::ios::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL); cout.precision(10); int main() { TLE; int t,n,m,k; cin>>t; while(t--) { cin>>n>>m>>k; if(n==m && n==k) cout<<n<<endl; else { lli ans = min(n,m); lli cnt = n+m+k - 3*ans; if(cnt>0) cout<<ans<<endl; else cout<<(3*ans+cnt)/3<<endl; } } return 0; }
A - Perfect Team CodeForces - 1221C
标签:iostream http cstring pre nsis and das row ==
原文地址:https://www.cnblogs.com/Shallow-dream/p/11618749.html