UVa 10465 Homer Simpson(DP完全背包)

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 霍默辛普森吃汉堡  有两种汉堡  一中吃一个需要m分钟  另一种吃一个需要n分钟  他共有t分钟时间 

要我们输出他在尽量用掉所有时间的前提下最多能吃多少个汉堡  如果时间无法用完  输出他吃的汉堡数和剩余喝酒的时间

很明显的完全背包问题  求两次  一次对个数  一次对时间就行了   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 10005using 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;}