zoukankan      html  css  js  c++  java
  • HDU 2126

    Buy the souvenirs

    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.
     
    #include<iostream>
    #include<cstdio>
    #include<cstring>
    #include<algorithm>
    #include<queue>
    using namespace std;
    int dp[505],p[35],cnt[505][35];
    int main()
    {
        int T;
        scanf("%d",&T);
        while(T--)
        {
            memset(dp,0,sizeof(dp));
            memset(cnt,0,sizeof(cnt));
            int n,m;
            scanf("%d%d",&n,&m);
            for(int i=1;i<=n;i++)
                scanf("%d",&p[i]);
            for(int i=1;i<=n;i++)
                for(int j=m;j>=p[i];j--)
                    dp[j]=max(dp[j],dp[j-p[i]]+1);
            if(dp[m]==0)
            {
                puts("Sorry, you can't buy anything.");
                continue;
            }
            cnt[0][0]=1;
            for(int i=1;i<=n;i++)
                for(int j=m;j>=p[i];j--)
                    for(int k=1;k<=35;k++)
                        cnt[j][k]+=cnt[j-p[i]][k-1];
            int ans=0;
            for(int i=m;i>=1;i--)
                if(dp[i]==dp[m])
                    ans+=cnt[i][dp[m]];
            printf("You have %d selection(s) to buy with %d kind(s) of souvenirs.
    ",ans,dp[m]);
        }
        return 0;
    }
  • 相关阅读:
    函数
    文件的基本操作
    c语言程序设计案例教程(第2版)笔记(一)—零散、输入输出、最小公倍数、选择排序、冒泡排序
    c语言中的rand()函数用法
    c语言 error C4996: 'strupr': The POSIX name for this item is deprecated. Instead, use the ISO C and C++ conformant name
    Python之列表生成式、生成器
    Python之迭代器
    Python之装饰器
    Linux之线程相关命令及常用命令
    重写、重构、重载区别
  • 原文地址:https://www.cnblogs.com/homura/p/6705624.html
Copyright © 2011-2022 走看看