Polycarp plays a well-known computer game (we won’t mention its name). In this game, he can craft tools of two types — shovels and swords. To craft a shovel, Polycarp spends two sticks and one diamond; to craft a sword, Polycarp spends two diamonds and one stick.
Each tool can be sold for exactly one emerald. How many emeralds can Polycarp earn, if he has a
sticks and b
diamonds?
Input
The first line contains one integer t
(1≤t≤1000) — the number of test cases.
The only line of each test case contains two integers a
and b (0≤a,b≤109) — the number of sticks and the number of diamonds, respectively.
Output
For each test case print one integer — the maximum number of emeralds Polycarp can earn.
Example Input
4
4 4
1000000000 0
7 15
8 7
Output
2
0
7
5
Note
In the first test case Polycarp can earn two emeralds as follows: craft one sword and one shovel.
In the second test case Polycarp does not have any diamonds, so he cannot craft anything.
思路：兩組方案，1.2a+b 2.a+2b，可以看出兩種方案的公共部分 a+b，假設m<n，用公共部分填滿m個，然後判斷n的剩餘部分（n-m）,如果剩餘部分過多（n-m>=m）,最終結果就是m（m<(m+n)/3），若剩餘部分不多（n-m<m）,那就先把剩餘部分填滿，再把多於部分充分利用，最終結果就是（m+n）/ 3，且此時（m+n）/ 3 < m，所以結果爲min( (m+n)/3，m）
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int mod = 1e9 + 7;
const int maxn = 1e6 + 1;
const int inf = 0x3f3f3f3f;
const double Pi = acos(-1.0);
const LL INF = 0x3f3f3f3f3f3f3f3f;
template<class T, class F> inline void mem(T a,F b, int c) {for(int i=0;i<=c;++i)a[i]=b;}
template<class T> inline void read(T &x,T xk=10) { // xk 为进制
char ch = getchar(); T f = 1, t = 0.1;
for(x=0; ch>'9'||ch<'0'; ch=getchar()) if(ch=='-')f=-1;
for(;ch<='9'&&ch>='0';ch=getchar())x=x*xk+ch-'0';if(ch=='.')
for(ch=getchar();ch<='9'&&ch>='0';ch=getchar(),t*=0.1)x+=t*(ch-'0');x*=f;
}
int main() {
int T; read(T);
while(T --) {
int n, m; read(n), read(m);
if (n < m) swap(n, m);
cout << min( (n + m) / 3, m ) << endl;
}
return 0;
}