UVa 10465 Homer Simpson

Return of the Aztecs

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

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Problem setter: Sadrul Habib Chowdhury
Solution author: Monirul Hasan (Tomal)

Time goes, you say? Ah no!
Alas, Time stays, we go.
-- Austin Dobson

 

有一我的喜欢吃burger，他吃每一个A-burger花费的时间为m，吃每一个B-burger花费的时间为n。求在t时间内，以浪费时间最小为前提，他最多能吃多少个burger

设dp[x]表示这我的花x时间（没有浪费）最多吃的burger个数，则有：

　　dp[i]=max{ dp[i-m], dp[i-n] }+1

最后从dp[t]倒着找，找到第一个离t最近的非0值即为答案

 

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<algorithm>
 5 
 6 using namespace std;
 7 
 8 int m,n,t;
 9 int dp[10050];
10 
11 int main()
12 {
13     while(scanf("%d %d %d",&m,&n,&t)==3)
14     {
15         memset(dp,-1,sizeof(dp));
16         dp[0]=0;
17 
18         for(int i=1;i<=t;i++)
19         {
20             if(i>=m&&dp[i-m]!=-1)
21                 dp[i]=dp[i-m]+1;
22             if(i>=n&&dp[i-n]!=-1)
23                 dp[i]=max(dp[i],dp[i-n]+1);
24         }
25 
26         int k=t;
27         while(dp[k]==-1)
28             k--;
29         printf("%d",dp[k]);
30         if(k!=t)
31             printf(" %d\n",t-k);
32         else
33             putchar('\n');
34     }
35 
36     return 0;
37 }

[C++]