zoukankan      html  css  js  c++  java
  • poj 1787 Charlie's Change

    Charlie's Change

    Time Limit: 1000MS   Memory Limit: 30000K

    Total Submissions: 5731   Accepted: 1849

    Description

    Charlie is a driver of Advanced Cargo Movement, Ltd. Charlie drives a lot and so he often buys coffee at coffee vending machines at motorests. Charlie hates change. That is basically the setup of your next task. 
    Your program will be given numbers and types of coins Charlie has and the coffee price. The coffee vending machines accept coins of values 1, 5, 10, and 25 cents. The program should output which coins Charlie has to use paying the coffee so that he uses as many coins as possible. Because Charlie really does not want any change back he wants to pay the price exactly. 

    Input

    Each line of the input contains five integer numbers separated by a single space describing one situation to solve. The first integer on the line P, 1 <= P <= 10 000, is the coffee price in cents. Next four integers, C1, C2, C3, C4, 0 <= Ci <= 10 000, are the numbers of cents, nickels (5 cents), dimes (10 cents), and quarters (25 cents) in Charlie's valet. The last line of the input contains five zeros and no output should be generated for it.

    Output

    For each situation, your program should output one line containing the string "Throw in T1 cents, T2 nickels, T3 dimes, and T4 quarters.", where T1, T2, T3, T4 are the numbers of coins of appropriate values Charlie should use to pay the coffee while using as many coins as possible. In the case Charlie does not possess enough change to pay the price of the coffee exactly, your program should output "Charlie cannot buy coffee.".

    Sample Input

    12 5 3 1 2 16 0 0 0 1 0 0 0 0 0 

    Sample Output

    Throw in 2 cents, 2 nickels, 0 dimes, and 0 quarters. 
    Charlie cannot buy coffee.

    题意:有1,5,10,25元硬币各C1, C2, C3, C4个,求正好可以买一杯价值P元的咖啡需要的最多的硬币数,输出每种硬币个数。

    题解:完全背包+记录路径。

    代码:
    #include<iostream>
    #include<algorithm>
    #include<cstdio>
    using namespace std;
    const int maxn=10005,inf=0x3f3f3f3f;
    int num[5],cnt[maxn],w[5]={0,1,5,10,25},dp[maxn],path[maxn],ans[30];
    int main()
    {
      int i,j,P;
      while(scanf("%d%d%d%d%d",&P,&num[1],&num[2],&num[3],&num[4]))
      {
       if(P+num[1]+num[2]+num[3]+num[4]==0) break;
       {
         fill(dp,dp+maxn,-inf);
         fill(path,path+maxn,-1);
         fill(ans,ans+30,0);
         dp[0]=0;
       }
       for(i=1;i<=4;i++)
       {
        fill(cnt,cnt+maxn,0);
        for(j=w[i];j<=P;j++)
         if(dp[j-w[i]]+1>dp[j]&&cnt[j-w[i]]+1<=num[i])
         {
          dp[j]=dp[j-w[i]]+1;
          cnt[j]=cnt[j-w[i]]+1;
          path[j]=w[i];
         }
       }
       if(dp[P]<0)  printf("Charlie cannot buy coffee.
    ");
       else
       {
         i=P;
         while(i>0)
           ans[path[i]]++,i-=path[i];
         printf("Throw in %d cents, %d nickels, %d dimes, and %d quarters.
    ",ans[1],ans[5],ans[10],ans[25]);
       }
      }
      return 0;
    }
    本博客仅为本人学习,总结,归纳,交流所用,若文章中存在错误或有不当之处,十分抱歉,劳烦指出,不胜感激!!!
  • 相关阅读:
    忍者必须死3 模拟器按键设置
    C# 工厂模式 个人基本流程
    WPF Boolean类型转化器收集 反转转化器
    Json实体类驼峰名称转化器
    TDengine + EMQ X + Grafana 轻松搭建高效低成本的边缘侧工业互联网平台
    呼声最高的数据更新功能来了,用户需要什么,我们就开源什么
    年轻人不讲武德,TDengine边缘侧数据存储方案挑战SQLite
    保姆级演示一分钟搞定TDengine的下载安装
    双汇大数据方案选型:从棘手的InfluxDB+Redis到毫秒级查询的TDengine
    HiveMQ TDengine extension 使用指南
  • 原文地址:https://www.cnblogs.com/VividBinGo/p/11366841.html
Copyright © 2011-2022 走看看