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 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
UVa 10465 Homer Simpson(DP 完全背包)
原文地址：http://blog.csdn.net/acvay/article/details/38776207
