分析： 我們假設購買一種可樂p瓶，我們可以得到期望：p*(m/n*a[i]+(n-m)/n*b[i])，由這個式子我們可以看出唯一的變量是i，所以可以遍歷i找出式子的最大值

#include <map>
#include <set>
#include <stack>
#include <cmath>
#include <queue>
#include <cstdio>
#include <vector>
#include <string>
#include <cstring>
#include 
 
<iostream>
#include <algorithm>
#define debug(a) cout << #a << " " << a << endl
using namespace std;
const int maxn = 2e5 + 10;
const int mod = 1e9 + 7;
typedef long long ll;
ll a[maxn], b[maxn];
int main() {
    ll n, m, k;
    while( cin >> n >> m >> k ) {
        ll sum 
 
= -1e9, j = 0, flag;
        for( ll i = 0; i < k; i ++ ) {
            cin >> a[i] >> b[i];
            if( m*a[i]+(n-m)*b[i] >= sum ) {
                flag = i;
                sum = m*a[i]+(n-m)*b[i];
            }
        }
        for( ll i = 0; i < k; i ++ ) {
            if
 
( m*a[i]+(n-m)*b[i] == sum && i == flag ) { //取字典序最小
                if( i == 0 ) {
                    cout << n;
                } else {
                    cout << " " << n;
                }
            } else {
                if( i == 0 ) {
                    cout << 0;
                } else {
                    cout << " " << 0;
                }
            }
        }
        cout << endl;
    }
    return 0;
}

CodeM 資格賽 B 可樂 思維