Python3 解法, 执行用时: 101ms, 内存消耗: 4744K, 提交时间: 2023-06-02 14:45:11

t=int(input())
for _ in range(t):
    [c1,c2, c3, c4]=list(map(int,input().split()))
    def f(c1,c2,c3,c4,n):
        c1-=4*n
        needc=n
        needh=n
        needd=n
        
        x=min(needh,c2)
        needh-=x
        c2-=x
        needc+=needh
        
        x=min(needc,c4)
        c4-=x
        needc-=x
        
        x=min(needc,c1,c3)
        c1-=x
        c3-=x
        needc-=x
        
        if c1<3*needc:
            return False
        c1-=3*needc
        
        x=min(c3,c2,needd)
        c3-=x
        c2-=x
        needd-=x
        
        x=min(c3,c4,needd)
        c3-=x
        c4-=x
        needd-=x
        
        x=min(c3//2,c1,needd)
        c3-=2*x
        c1-=x
        needd-=x
        
        x=min(c1//2,c2,needd)
        c1-=2*x
        c2-=x
        needd-=x
         
        x=min(c1//2,c4,needd)
        c1-=2*x
        c4-=x
        needd-=x
         
        x=min(c1//3,c3,needd)
        c1-=3*x
        c3-=x
        needd-=x
        
        if c1<5*needd:
            return False
        return True
        
    l=1
    r=c1//4
    while l<=r:
        mid=(l+r)>>1
        if f(c1,c2,c3,c4,mid):
            l=mid+1
        else:
            r=mid-1
    print(r)

C++(g++ 7.5.0) 解法, 执行用时: 7ms, 内存消耗: 448K, 提交时间: 2023-06-02 15:14:56

#include <bits/stdc++.h>

using namespace std;
 
int a, b, c, d;
 
bool check (int s) {
    int _a = a, _b = b, _c = c, _d = d;
    _a -= 3 * s;
    int p1 = s, p2 = s, p3 = s;
    if (_d <= s)
        p3 -= _d;
    else
        p3 = 0, _b += _d - s;
    if (_b <= s) 
        p2 -= _b;
    else
        p2 -= s, p1 = max(0, p1 - _b + s);
    int O = (s - p1) + p1 * 2 + p2 + p3;
    int P = (s - p1) * 2 + p1 * 5 + (s - p2) * 1 + p2 * 4 + p3 * 3;
    O = min(O, _c);
    P -= O * 2;
    return _a - P >= 0;
}
 
int main() {
    int T;
    cin >> T;
    while (T --) {
        cin >> a >> b >> c >> d;
        int l = 0, r = 1e4 + 10, m;
        while (l < r - 1) {
            m = (l + r) >> 1;
            if (check(m))
                l = m;
            else
                r = m;
        }
        printf("%d\n", l);
    }
    return 0;
}

C++(clang++ 11.0.1) 解法, 执行用时: 7ms, 内存消耗: 508K, 提交时间: 2023-06-27 23:52:36

#include<iostream>
using namespace std;
int a,b,c,d;
bool check(int x){
    int _a = a,_b = b,_c = c,_d = d;
    _a -= 3 * x;
    _d -= x;
    if(_d < 0) {
        _c += _d;
        _a += _d;
        _d = 0;
    }
    if(_b + _d < 2 * x){
        _c -= 2*x -(_b + _d);
        _a -= 2*x -(_b + _d);
    }
    _c -= x;
    if(_c < 0) _a+=_c*2;
    _a -= x;
    if(_a < 0) return false;
    else return true;
}
int main(){
    int t;
    cin>>t;
    while(t--){
        cin>>a>>b>>c>>d;
        int l = 0;
        int r = 10005;
        int mid;
        while(l < r){
            mid = (l + r + 1)>>1;
            if(check(mid)) l = mid;
            else r = mid - 1;
        }
        cout<<l<<endl;
    }

    return 0;
}