Homer Simpson - UVa 10465 dp

Problem C:	Homer Simpson
Time Limit: 3 seconds Memory Limit: 32 MB

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Homer Simpson, a very smart guy, likes eating Krusty-burgers. It takes Homer m minutes to eat a Krusty- burger. However, there�s a new type of burger in Apu�s Kwik-e-Mart. Homer likes those too. It takes him n minutes to eat one of these burgers. Given t minutes, you have to find out the maximum number of burgers Homer can eat without wasting any time. If he must waste time, he can have beer.

Input

Input consists of several test cases. Each test case consists of three integers m, n, t (0 < m,n,t < 10000). Input is terminated by EOF.

Output

For each test case, print in a single line the maximum number of burgers Homer can eat without having beer. If homer must have beer, then also print the time he gets for drinking, separated by a single space. It is preferable that Homer drinks as little beer as possible.

Sample Input

3 5 54
3 5 55

Sample Output

18
17

题意：如果能正好在规定时间一直在吃汉堡，就输出最多吃多少汉堡，否则输出最少喝多少时间的啤酒的情况下，最多吃多少汉堡。

AC代码如下：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int dp[10010];
int main()
{ int n,m,t,i,j,k;
  while(~scanf("%d%d%d",&n,&m,&t))
  { memset(dp,-1,sizeof(dp));
    dp[0]=0;
    for(i=n;i<=t;i++)
     if(dp[i-n]>=0)
      dp[i]=dp[i-n]+1;
    for(i=m;i<=t;i++)
     if(dp[i-m]>=0)
     { if(dp[i]==-1)
        dp[i]=dp[i-m]+1;
       else
        dp[i]=max(dp[i],dp[i-m]+1);
     }
    if(dp[t]>=0)
     printf("%d\n",dp[t]);
    else
    { for(k=t;k>=0;k--)
       if(dp[k]>=0)
        break;
      printf("%d %d\n",dp[k],t-k);
    }
  }
}