Time Limit: 10000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 2393 Accepted Submission(s): 914
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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//有n个物品每个物品的价格是v[i],现在有m元钱问最多买多少种物品，并求出有多少种选择方法； //如果将物品的价格看做容量，将它的件数1看做价值的话，那么用01背包就可以求的花费m钱所能买到的最大件数dp[m]。 //但是题目还要求方案数，因此很容易想到再建立一个数组p[j]，存储j元钱能买dp[j]个物品的方案数。在求解01背包的过程中，要分两种情况讨论： #include<stdio.h> #include<cstring> #define N 505 int main () { int t, n, m, val[N], dp[N], p[N]; scanf ("%d", &t); while (t--) { memset(dp, 0,sizeof(dp)); //初始化 memset(p,0,sizeof(p)); //初始化 scanf ("%d %d", &n, &m); for (int i=1; i<=n; i++) scanf ("%d", &val[i]); for (int i=1; i<=n; i++) for (int j=m; j>=val[i]; j--) { //若选了物品i后，能买的件数比不选物品i的件数大，即dp[j-val[i]]>dp[j]那么 //更新dp[j]，同时，p[j]的方案数即为p[j-val[i]] //例如 AB AC ABD ACD 仍为两种，所以p[j]=p[j-val[i]]即可 if (dp[j] < dp[j-val[i]]+1) { dp[j] = dp[j-val[i]]+1; //要注意p[j-val[i]]为0的情况，因此当它为0时，p[j]=1，1即为D if (!p[j-val[i]]) p[j-val[i]]++; p[j] = p[j-val[i]]; } //若选了物品i后，能买的件数与不选物品i的件数相同， else if (dp[j] == dp[j-val[i]]+1) { if (!p[j-val[i]]) p[j-val[i]]++; //也要注意p[j-val[i]]=0的情况，当它为0时，p[j]+=1，1即为D p[j] += p[j-val[i]]; //总的方案数即为p[j]+p[j-val[i]] } } if (!dp[m]) puts ("Sorry, you can't buy anything."); else printf ("You have %d selection(s) to buy with %d kind(s) of souvenirs.\n", p[m], dp[m]); } return 0; }
