hdu 4091 Zombie’s Treasure Chest

需要仔细体会s1和s2都很小的时候，其枚举量为什么是s1+s2

#include<cstdio>
#include<algorithm>
#include<iostream>
using namespace std;
int t,kase;
int main()
{
    scanf("%d",&t);
    kase=1;
    while(t--)
    {
        long long n,s1,v1,s2,v2;
        long long ans=0;
        cin>>n>>s1>>v1>>s2>>v2;
        printf("Case #%d: ",kase++);
        if(s1 > s2)//体积小的去前边
        {
            swap(s1, s2);
            swap(v1, v2);
        }
        if(n / s2 >= 65536)
        {
            for(long long  i = 0; i <= s1; i++)
            {
                ans = max(ans, v2*i + (n-s2*i)/s1*v1);
            }
            for(long long i = 0; i <= s2; i++)
            {
                ans = max(ans, v1*i + (n-s1*i)/s2*v2);
            }
        }
        else   // s2 is large
        {
            for(long long i = 0; s2*i <= n; i++)
                ans = max(ans, v2*i + (n-s2*i)/s1*v1);
        }
        printf("%lld\n",ans);
    }
    return 0;
}