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).

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.

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.”

wangye

 

题意： n个物品对应不同的价值且只有一件   现在有m元  问 在最多能购买k种物品的情况下 有s 种购买方案。

输出 You have s selection(s) to buy with k kind(s) of souvenirs.

否则输出 Sorry, you can‘t buy anything.

 

题解：  不要写空行 有空行pe  坑

           01背包 增加一维  f[j][k] 表示 j元钱 购买k种（件）物品有多少种方案

           方程 f[j][k]=f[j][k]+f[j-val[i]][k-1];

 1 #include<iostream>
 2 #include<cstring>
 3 #include<cstdio>
 4 #include<algorithm>
 5 #define ll __int64
 6 using namespace std;
 7 int t;
 8 int n,m;
 9 int val[35];
10 int f[505][35];
11 int flag;
12 int main() 
13 {
14    while(scanf("%d",&t)!=EOF)
15    {
16        for(int i=1;i<=t;i++)
17        {
18            scanf("%d %d",&n,&m);
19         memset(f,0,sizeof(f));
20            memset(val,0,sizeof(val));
21         for(int j=0;j<=m;j++)
22             f[j][0]=1;
23         flag=-1;
24            for(int j=1;j<=n;j++)
25             scanf("%d",&val[j]);
26            for(int j=1;j<=n;j++)
27             for(int k=m;k>=val[j];k--)
28                   {
29                       for(int g=j;g>=1;g--)
30                 {
31                 f[k][g]=f[k][g]+f[k-val[j]][g-1];
32                 if(f[k][g])
33                  {
34                      if(flag<g)
35                       flag=g;
36                  }
37                 }
38             }    
39        if(flag==-1)
40        printf("Sorry, you can‘t buy anything.\n");
41        else
42        printf("You have %d selection(s) to buy with %d kind(s) of souvenirs.\n",f[m][flag],flag);
43      }
44    }    
45     return 0;
46 }