C. Om Nom and Candies
Time Limit: 1 Sec Memory Limit: 256 MB
题目连接
http://codeforces.com/contest/526/problem/C
Description
A sweet little monster Om Nom loves candies very much. One day he found himself in a rather tricky situation that required him to think a bit in order to enjoy candies the most. Would you succeed with the same task if you were on his place?
One day, when he came to his friend Evan, Om Nom didn‘t find him at home but he found two bags with candies. The first was full of blue candies and the second bag was full of red candies. Om Nom knows that each red candy weighs Wr grams and each blue candy weighs Wb grams. Eating a single red candy gives Om Nom Hr joy units and eating a single blue candy gives Om Nom Hb joy units.
Candies are the most important thing in the world, but on the other hand overeating is not good. Om Nom knows if he eats more than C grams of candies, he will get sick. Om Nom thinks that it isn‘t proper to leave candy leftovers, so he can only eat a whole candy. Om Nom is a great mathematician and he quickly determined how many candies of what type he should eat in order to get the maximum number of joy units. Can you repeat his achievement? You can assume that each bag contains more candies that Om Nom can eat.
Input
The single line contains five integers C, Hr, Hb, Wr, Wb (1 ≤ C, Hr, Hb, Wr, Wb ≤ 109).
Output
Print a single integer — the maximum number of joy units that Om Nom can get.
Sample Input
Sample Output
HINT
题意
只有两个物品的多重背包问题
题解：
啊， 正解实在不会，只能乱搞
//qscqesze
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define maxn 200001
#define mod 10007
#define eps 1e-9
//const int inf=0x7fffffff; //无限大
const int inf=0x3f3f3f3f;
/*
int buf[10];
inline void write(int i) {
int p = 0;if(i == 0) p++;
else while(i) {buf[p++] = i % 10;i /= 10;}
for(int j = p-1; j >=0; j--) putchar(‘0‘ + buf[j]);
printf("\n");
}
*/
//**************************************************************************************
inline ll read()
{
int x=0,f=1;char ch=getchar();
while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}
return x*f;
}
int main()
{
ll c,hr,hb,wr,wb;
ll ans=0;
cin>>c>>hr>>hb>>wr>>wb;
if(wr==wb)
{
cout<<c/wr*max(hr,hb)<<endl;
return 0;
}
if(wr<wb)
{
swap(wr,wb);
swap(hr,hb);
}
int time=0;
for(int i=c/wr;i>=0;i--)
{
time++;
ll sum=i*hr+(c-wr*i)/wb*hb;
ans=max(sum,ans);
if(time>10000000)
break;
}
//cout<<ans<<endl;
swap(wr,wb);
swap(hr,hb);
time=0;
for(int i=c/wr;i>=0;i--)
{
time++;
ll sum=i*hr+(c-wr*i)/wb*hb;
ans=max(sum,ans);
if(time>10000000)
break;
}
cout<<ans<<endl;
}
