Buy the souvenirs

Time Limit: 10000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1662    Accepted Submission(s):
611

Problem Description

When the winter holiday comes, a lot of people will
have a trip. Generally, there are a lot of souvenirs to sell, and sometimes the
travelers will buy some ones with pleasure. Not only can they give the souvenirs
to their friends and families as gifts, but also can the souvenirs leave them
good recollections. All in all, the prices of souvenirs are not very dear, and
the souvenirs are also very lovable and interesting. But the money the people
have is under the control. They can’t buy a lot, but only a few. So after they
admire all the souvenirs, they decide to buy some ones, and they have many
combinations to select, but there are no two ones with the same kind in any
combination. Now there is a blank written by the names and prices of the
souvenirs, as a top coder all around the world, you should calculate how many
selections you have, and any selection owns the most kinds of different
souvenirs. For instance:

And you have only 7
RMB, this time you can select any combination with 3 kinds of souvenirs at most,
so the selections of 3 kinds of souvenirs are ABC (6), ABD (7). But if you have
8 RMB, the selections with the most kinds of souvenirs are ABC (6), ABD (7), ACD
(8), and if you have 10 RMB, there is only one selection with the most kinds of
souvenirs to you: ABCD (10).

Input

For the first line, there is a T means the number
cases, then T cases follow.
In each case, in the first line there are two
integer n and m, n is the number of the souvenirs and m is the money you have.
The second line contains n integers; each integer describes a kind of souvenir.

All the numbers and results are in the range of 32-signed integer, and
0<=m<=500, 0<n<=30, t<=500, and the prices are all positive
integers. There is a blank line between two cases.

Output

If you can buy some souvenirs, you should print the
result with the same formation as “You have S selection(s) to buy with K kind(s)
of souvenirs”, where the K means the most kinds of souvenirs you can buy, and S
means the numbers of the combinations you can buy with the K kinds of souvenirs
combination. But sometimes you can buy nothing, so you must print the result
“Sorry, you can‘t buy anything.”

Sample Input

2

4 7

1 2 3 4

4 0

1 2 3 4

Sample Output

You have 2 selection(s) to buy with 3 kind(s) of souvenirs.

Sorry, you can‘t buy anything.

Author

wangye

Source

HDU
2007-10 Programming Contest

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记录方案总数的01背包，给dp数组加一维，记录方案总数，记住需要初始化，把所有 dp[i][1] 都设置成1。

题意：n个物品，m元钱，每个物品最多买一次，问最多可以买几件物品，并且输出方案数。加一维表示已经买几件物品。

附上代码：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int main()
{
    int T,n,m,i,j;
    int a[35],dp[505][2];
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        for(i=0; i<n; i++)
            scanf("%d",&a[i]);
        memset(dp,0,sizeof(dp));
        for(i=0; i<=m; i++) dp[i][1]=1;  //先全部标记为1，能拿这么多件物品，起码有一种方案
        for(i=0; i<n; i++)
            for(j=m; j>=a[i]; j--)
            {
                if(dp[j][0]==dp[j-a[i]][0]+1)   //如果记录了j元钱拿的物品数量，正好与此时记录的不拿这件物品的最多物品加1相等
                    dp[j][1]=dp[j-a[i]][1]+dp[j][1];
                else if(dp[j][0]<dp[j-a[i]][0]+1)  //如果小于，则更新dp
                {
                    dp[j][0]=dp[j-a[i]][0]+1;
                    dp[j][1]=dp[j-a[i]][1];
                }
            }
        if(dp[m][0]!=0)
            printf("You have %d selection(s) to buy with %d kind(s) of souvenirs.\n",dp[m][1],dp[m][0]);
        else
            printf("Sorry, you can‘t buy anything.\n");
    }
    return 0;
}