码题集OJ-补给 (matiji.net)

思想

我们每次去枚举每个数打折扣的情况

code

#include <bits/stdc++.h>
#define int long long
using namespace std;
int N, B;
const int M = 1010;
int p[M], s[M], t[M];
int ans;

signed main() {
    cin.tie(nullptr)->sync_with_stdio(false);
	cin >> N >> B;
	for (int i = 0; i < N; i++)
		cin >> p[i] >> s[i];

	//枚举每个优惠的方案
	for (int i = 0; i < N; i++) {
		t[i] = p[i] / 2 + s[i];
		
		for (int j = 0; j < N; j++) {
			if (i == j)continue;
			else t[j] = p[j] + s[j];

		}

		sort(t, t + N);
		for (int k = 0; k < N; k++) {
			if (B < t[k]) {
				ans = max(ans, k);
				break;
			} else {
				B -= t[k];
			}
		}
	}
	cout << ans << endl;
	return 0;
}