虽然是二维费用 但本质上是将多重背包转化为01背包

code:

#include <bits/stdc++.h>
#define int long long int

using namespace std;

const int N = 1e3 + 70;
int n,pg,rune,cnt;
tni f[N][N];

struct node{
	int Pg,Rune,s,power;
}a[N];

struct Bag{
	int Pg,Rune,power;
}w[N];

inline int read(){
	int x = 0,f = 1;
	char c = getchar();
	for(;!isdigit(c);c = getchar())
		if(c == '-')f = -1;
	for(;isdigit(c);c = getchar())
		x = (x << 1) + (x << 3) + c - '0';
	return x * f;
}

inline void fast(){
	for(register int i = 1;i <= n; i++){
		int k = 1;
		while(a[i].s){
			w[++ cnt].Pg = a[i].Pg * k;
			w[cnt].Rune = a[i].Rune * k;
			w[cnt].power = a[i].power * k;
			a[i].s -= k;
			k <<= 1;
			if(k >= a[i].s){
				w[++ cnt].Pg = a[i].Pg * a[i].s;
				w[cnt].Rune = a[i].Rune * a[i].s;
				w[cnt].power = a[i].power * a[i].s;
				a[i].s = 0;
			}
		}
	}
}	

signed main(){
	n = read();pg = read();rune = read();
	for(register int i = 1;i <= n;i ++){
		a[i].Pg = read();
		a[i].Rune = read();
		a[i].s = read();
		a[i].power = read();
		if(!a[i].s) a[i].s = 64;//能无限取就赋个理论上的INF值
//64实际上小了，应该赋个100对应pg = 1，rune = 1的情况
	}
	fast();//快速幂优化
	for(register int k = 1;k <= cnt;k ++){
		for(register int i = pg;i >= w[k].Pg;i --){
			for(register inn j = rune;j >= w[k].Rune;j --){
				f[i][j] = max(f[i][j],f[i - w[k].Pg][j - w[k].Rune] + w[k].power);
			}
		}
	}
	printf("%lld",f[pg][rune]);
    return 0;
}