UVA 10465 —— 完全背包

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本文探讨了一个经典的算法问题：如何在有限时间内最大化食用两种不同汉堡的数量。通过使用动态规划的方法，解决了一个具体实例的问题，并提供了完整的代码实现。

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10465 - Homer Simpson

Time limit: 3.000 seconds

Return of the Aztecs

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

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

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

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

题意是一个人要在时间t内吃汉堡，有两种汉堡，需要的时间分别为m和n，问你最多能吃多少个，如果其中有间隔输出间隔时间。

思路：只有两件物品的完全背包。

/*
ID: xinming2
PROG: stall4
LANG: C++
*/
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <cstring>
#include <map>
#include <string>
#include <stack>
#include <cctype>
#include <vector>
#include <queue>
#include <set>
#include <utility>
#include <cassert>
using namespace std;
///#define Online_Judge
#define outstars cout << "***********************" << endl;
#define clr(a,b) memset(a,b,sizeof(a))
#define lson l , mid  , rt << 1
#define rson mid + 1 , r , rt << 1 | 1
#define mk make_pair
#define FOR(i , x , n) for(int i = (x) ; i < (n) ; i++)
#define FORR(i , x , n) for(int i = (x) ; i <= (n) ; i++)
#define REP(i , x , n) for(int i = (x) ; i > (n) ; i--)
#define REPP(i ,x , n) for(int i = (x) ; i >= (n) ; i--)
const int MAXN = 10000 + 50;
const int sigma_size = 26;
const long long LLMAX = 0x7fffffffffffffffLL;
const long long LLMIN = 0x8000000000000000LL;
const int INF = 0x7fffffff;
const int IMIN = 0x80000000;
#define eps 1e-8
const int mod = (int)1e9 + 7;
typedef long long LL;
const LL MOD = 1000000007LL;
const double PI = acos(-1.0);

typedef pair<int , int> pi;
#define Bug(s) cout << "s = " << s << endl;
///#pragma comment(linker, "/STACK:102400000,102400000")
int n , m , t;
int dp[MAXN];
int main()
{

    while(~scanf("%d%d%d" , &m , &n, &t ))
    {
        clr(dp , -1);
        dp[0] = 0;
        for(int i = m ; i <= t ; i++)
        {
            if(dp[i - m] != -1)dp[i] = max(dp[i] , dp[i - m] + 1);
        }
        for(int i = n ; i <= t  ; i++)
        {
            if(dp[i - n] != -1)dp[i] = max(dp[i] , dp[i - n] + 1);
        }
        if(dp[t] != -1)printf("%d\n" , dp[t]);
        else
        {
            for(int i = t ; i >= 0 ; i--)
            {
                if(dp[i] != -1)
                {
                    printf("%d %d\n" , dp[i], t - i);
                    break;
                }
            }
        }
    }
    return 0;
}