链接：https://ac.nowcoder.com/acm/contest/4743/C
换元后三分，这题三分是利用三分确定极值存在的一个区间，然后在这个区间上再求解。

#include <bits/stdc++.h>
using namespace std;
#define IO ios::sync_with_stdio(false);cin.tie(0); cout.tie(0)
#define all(__vv__) (__vv__).begin(), (__vv__).end()
#define endl "\n"
#define SZ(x) ((int)(x).size())
#define pb push_back
#define pii pair<int, int>
#define mset(__x__,__val__) memset(__x__, __val__, sizeof(__x__))
typedef long long ll; typedef unsigned long long ull; typedef long double ld;
inline ll read() { ll s = 0, w = 1; char ch = getchar(); for (; !isdigit(ch); ch = getchar()) if (ch == '-') w = -1; for (; isdigit(ch); ch = getchar())    s = (s << 1) + (s << 3) + (ch ^ 48); return s * w; }
inline void print(ll x, int op = 10) { if (!x) { putchar('0'); if (op)    putchar(op); return; }    char F[40]; ll tmp = x > 0 ? x : -x;    if (x < 0)putchar('-');    int cnt = 0;    while (tmp > 0) { F[cnt++] = tmp % 10 + '0';        tmp /= 10; }    while (cnt > 0)putchar(F[--cnt]);    if (op)    putchar(op); }
inline ll gcd(ll x, ll y) { return y ? gcd(y, x % y) : x; }
ll qpow(ll a, ll b) { ll ans = 1;    while (b) { if (b & 1)    ans *= a;        b >>= 1;        a *= a; }    return ans; }    ll qpow(ll a, ll b, ll mod) { ll ans = 1; while (b) { if (b & 1)(ans *= a) %= mod; b >>= 1; (a *= a) %= mod; }return ans % mod; }
const int dir[][2] = { {0,1},{1,0},{0,-1},{-1,0},{1,1},{1,-1},{-1,1},{-1,-1} };
const int MOD = 1e9+7;
const int INF = 0x3f3f3f3f;
const int N = 5e5+10;
const double eps = 1e-11;
int a[N], b[N], c[N];
int t, x, y;

inline int cal(int m)
{
    return m + min((x-2*m)/4, (y-3*m));
}

int main()
{
    t = read();
    while(t--)
    {
        x = read(), y = read();
        int l = 0, r = min(x/2, y/3);
        while(r-l >= 5)
        {
            int Lmid = l + (r-l)/3;
            int Rmid = r - (r-l)/3;
            if(cal(Lmid) < cal(Rmid))
                l = Lmid;
            else 
                r = Rmid;
        }
        int maxx =  0;
        for(int i = l; i <= r; i++){
            maxx = max(maxx, cal(i));
        }
        print(maxx);
    }
}