UVa 10465 Homer Simpson(DP 全然背包)

Posted 2020-09-26 gccbuaa



题意 霍默辛普森吃汉堡  有两种汉堡  一中吃一个须要m分钟  还有一种吃一个须要n分钟  他共同拥有t分钟时间    要我们输出他在尽量用掉全部时间的前提下最多能吃多少个汉堡  假设时间无法用完  输出他吃的汉堡数和剩余喝酒的时间

非常明显的全然背包问题  求两次  一次对个数  一次对时间即可了   时间用不完的情况下就输出时间的

d1为个数的  d2为时间的  dt保存时间

#include<cstdio>
#include<cstring>
#include<algorithm>
#define maxn 10005
using namespace std;
int w[2],t,d1[maxn],d2[maxn],dt[maxn];
int main()
{
    while (scanf("%d%d%d",&w[0],&w[1],&t)!=EOF)
    {
        memset(d1,0x8f,sizeof(d1));
        memset(dt,0,sizeof(dt));
        memset(d2,0,sizeof(d2));
        d1[0]=0;
        for(int i=0; i<2; ++i)
            for(int j=w[i]; j<=t; ++j)
            {
                d1[j]=max(d1[j],d1[j-w[i]]+1);
                if((dt[j]<dt[j-w[i]]+w[i])||((dt[j]==dt[j-w[i]]+w[i])&&(d2[j]<d2[j-w[i]]+1)))
                {
                    dt[j]=dt[j-w[i]]+w[i];
                    d2[j]=d2[j-w[i]]+1;
                }
            }
        if(dt[t]==t)
            printf("%d\n",d1[t]);
        else
            printf("%d %d\n",d2[t],t-dt[t]);
    }
    return 0;
}

Homer Simpson

---

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