[codeforces | greedy]B. RPG Protagonist

这题主要在于不能一次性的求出贪心结果，必须遍历第一个角色然后求出第二个角色的贪心结果。
时间复杂度为O(n)

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define debug(x) cout<<#x<<" is "<<x<<endl
ll ans;
int main(){
    
    
	int t;
	//freopen("input.txt","r",stdin);
	scanf("%d",&t);
	while(t--){
    
    
		ll p,f,cnts,cntw,s,w;
		ans = 0;
		scanf("%lld%lld%lld%lld%lld%lld",&p,&f,&cnts,&cntw,&s,&w);
		if(p < f)swap(p,f);
		if(s > w){
    
    
			swap(s,w);
			swap(cnts,cntw);
		}
		for(int i = 0;i <= cnts;i++){
    
    
			if(i * s > p)break;
			ll tcnts = cnts,tcntw = cntw,tp = p,tf = f;
			ll consum_w = min(tcntw,(tp - i * s) / w);
			ll t_ans = i + consum_w;
			tcnts -= i;
			tcntw -= consum_w;
			ll f_consum_s = min(tcnts,tf / s);
			ll f_consum_w = min(tcntw,(tf - f_consum_s * s) / w);
			t_ans += f_consum_s + f_consum_w;
			ans = max(ans,t_ans);
		}
		cout<<ans<<endl;
	}
	return 0;
}