B. RPG Protagonist[Educational Codeforces Round 94 (Rated for Div. 2)]数学枚举

B. RPG Protagonist

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题目大意：就是你有两个人，每个人都有一个最大的体力值p和f，这两个人要去搬运剑和盾牌，剑的数量是cnts,盾的数量是cntw,每个剑的重量是是s，每个盾的重量是w，问你最多可以搬运多少武器

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解题思路：一眼就是一个贪心题但是如何贪心确是一个问题？
很显然剑和盾哪个轻就拿哪个为了直观一点，我们可以设出这些自变量了

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#include <iostream>
#include <cstdio>
#include <stack>
#include <sstream>
#include <limits.h>
#include <vector>
#include <map>
#include <cstring>
#include <deque>
#include <cmath>
#include <iomanip>
#include <unordered_map>
#include <queue>
#include <algorithm>
#include <set>
#define mid ((l + r) >> 1) 
#define Lson rt << 1, l , mid
#define Rson rt << 1|1, mid + 1, r
#define ms(a,al) memset(a,al,sizeof(a))
#define log2(a) log(a)/log(2)
#define _for(i,a,b) for( int i = (a); i < (b); ++i)
#define _rep(i,a,b) for( int i = (a); i <= (b); ++i)
#define for_(i,a,b) for( int i = (a); i >= (b); -- i)
#define rep_(i,a,b) for( int i = (a); i > (b); -- i)
#define lowbit(x) ((-x) & x)
#define IOS std::ios::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define INF 0x3f3f3f3f
#define LLF 0x3f3f3f3f3f3f3f3f
#define hash Hash
#define next Next
#define pb push_back
#define f first
#define s second
using namespace std;
const int N = 1e7 + 10, MOD = 1e9 + 7;
const int maxn = 2e5;
const long double eps = 1e-5;
const int EPS = 500 * 500;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PII;
typedef pair<ll,ll> PLL;
typedef pair<double,double> PDD;
template<typename T> void read(T &x)
{
    x = 0;char ch = getchar();ll f = 1;
    while(!isdigit(ch)){if(ch == '-')f*=-1;ch=getchar();}
    while(isdigit(ch)){x = x*10+ch-48;ch=getchar();}x*=f;
}
template<typename T, typename... Args> void read(T &first, Args& ... args) 
{
    read(first);
    read(args...);
}
int T, n;
string s;
int main()
{
    read(T);
    while(T --)
    {
        int p, f;
        int cs,cw;
        int s, w;
        read(p,f);
        read(cs,cw);
        read(s,w);
        if (s>w){swap(s,w);swap(cs,cw);}
        int ans = 0;
        for (int s1=0;s1*s<=p && s1<=cs;s1++){
			int w1=min((p-s1*s)/w,cw);
			int s2=min(cs-s1,f/s);
			int w2=min((f-s2*s)/w,cw-w1);
			ans=max(ans,s1+s2+w1+w2);
		}
		cout << ans << endl;
    }
    return 0;
}