題目戳這裡
對於三種禮物來說，都需要用到一個a、一個b，所以抽象來說每個禮物對應a、b、c。於是二分答案，見程式碼

#include <iostream>
#include <algorithm>
#include <queue>
#include <stack>
#include <cstdio>
#include <string>
#include <cstring>
#include <vector>
#include <set>
#define LL long long
 

#define INF 0x3f3f3f3f
#define MOD 998244353
const int maxn = 1e5 + 5;
using namespace std;
bool judge(int a, int b, int c, int x){
    a -= x;
    b -= x;
    if(a < 0 || b < 0) return false;
    if(a + b + c < x) return false;//保證了有x份
    if(a > b) swap(a,b);
    if(b > c) swap(b,c);
    if
 
(a + b < x / 2) return false;//保證不相鄰
    return true;
}
int main(){
    int T;
    scanf("%d",&T);
    while(T--){
        int a,b,c;
        scanf("%d%d%d",&a,&b,&c);
        int l = 0, r = min(a,b);
        int ans = 0, m;
        while(l <= r){
            m = (l + r) / 2;
            if
 
(judge(a,b,c,m)){
                ans = m;
                l = m + 1;
            }
            else r = m - 1;
        }
        printf("%d\n",ans);
    }
}
/*
 2
 4 4 0
 1 1 1
 */