4393: [Usaco2015]Fruit Feast
Time Limit: 10 Sec Memory Limit: 128 MBSubmit: 81 Solved: 50
[Submit][Status][Discuss]
Description
Bessie has broken into Farmer John's house again! She has discovered a pile of lemons and a pile of oranges in the kitchen (effectively an unlimited number of each), and she is determined to eat as much
as possible.
Bessie has a maximum fullness of T (1≤T≤5,000,000). Eating an orange increases her fullness by A, and eating a lemon increases her fullness by B (1≤A,B≤T). Additionally, if she wants, Bessie can drink water at most one time, which will instantly decrease her
fullness by half (and will round down).
Help Bessie determine the maximum fullness she can achieve!
奶牛Bessie潜入了农夫约翰的家，她发现这里有无穷无尽的柠檬派和橘子派。
Bessie的饱胀值一開始是0。且上限是T，每一个柠檬派能够提供A点饱胀值。每一个橘子派能够提供B点饱胀值。
Bessie能够不断地吃东西，假设她的饱胀值没有超出T的话。同一时候，Bessie有一次喝水的机会，喝完后，她的饱胀值将降低一半（往下取整）。
请计算出Bessie的饱胀值最多能够达到多少。
Input
The first (and only) line has three integers T, A, and B.
Output
A single integer, representing the maximum fullness Bessie can achieve.
Sample Input
Sample Output
HINT
Source
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<algorithm>
#include<queue>
#define F(i,j,n) for(int i=j;i<=n;i++)
#define D(i,j,n) for(int i=j;i>=n;i--)
#define ll long long
#define pa pair<int,int>
#define maxn 5000100
#define inf 1000000000
using namespace std;
int a,b,t,ans;
bool f[maxn][2];
inline int 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()
{
t=read();a=read();b=read();
f[0][0]=true;
F(j,0,1) F(i,0,t) if (f[i][j])
{
if (i+a<=t) f[i+a][j]=true;
if (i+b<=t) f[i+b][j]=true;
if (!j) f[i/2][1]=true;
}
ans=t;
while (!f[ans][0]&&!f[ans][1]) ans--;
printf("%d
",ans);
}