普通数学解

#include <iostream>
using namespace std;

int n, m, a, b;

int main () {

    cin >> n >> m;
    cin >> a >> b;
    cout << (n + m) / (a + b) << endl;

    return 0;
}

但肯定不应是正解

暴力枚举法

枚举方案1

#include <iostream>
using namespace std;

#define int long long
int n, m, a, b;

signed main () {

    cin >> n >> m;
    cin >> a >> b;
    int ans = 0, leva = n / a; 
    for (int i = 0; i <= leva; i ++) {
        int lev_n = n - i * a;
        int lev_m = m - i * b;
        if (lev_m < 0) break;
        int pik_j = min(lev_n / b, lev_m / a);
        ans = max(ans, pik_j + i);
    }
    cout << ans << endl;

    return 0;
}

并未考虑 n 极大，a 极小数据，超时应是必然

暴力简单优化

优化特判 a == b 的情况，并且显然是单峰情况，一旦下降，则有解。同样存在极限超时情况

#include <iostream>
using namespace std;

int n, m, a, b;

int main () {

    cin >> n >> m;
    cin >> a >> b;
    if (a == b) {
        cout << min(n / a, m / b) << endl;
        return 0;
    }
    int ans = 0, leva = n / a; 
    for (int i = 0; i <= leva; i ++) {
        int lev_n = n - i * a;
        int lev_m = m - i * b;
        if (lev_m < 0) break;
        int pik_j = min(lev_n / b, lev_m / a);
        if (pik_j + i >= ans) {
            ans = pik_j + i;
        } else break;
    }
    cout << ans << endl;

    return 0;
}

暴力"玄学"优化

简单来说，在超时的极限赌一把，能过就不再优化了

#include <iostream>
using namespace std;

int n, m, a, b;

int main () {

    cin >> n >> m;
    cin >> a >> b;
    if (a == b) {
        cout << min(n / a, m / b) << endl;
        return 0;
    }
    if (n > m) swap(n, m);
    if (a > b) swap(a, b);
    
    int ans = 0, leva = min(n / a, m / a); 
    for (int i = 0; i <= leva; i ++) {
        if (i >= 5e7) {
            ans = max((n + m) / (a + b), ans);
            break;
        }
        int lev_n = n - i * a;
        int lev_m = m - i * b;
        if (lev_m < 0) break;
        int pik_j = min(lev_n / b, lev_m / a);
        if (pik_j + i >= ans) {
            ans = pik_j + i;
        } else break;
    }
    cout << ans << endl;

    return 0;
}

很好，用数学法赌过了，不再优化