zoukankan      html  css  js  c++  java
  • uva674

    题目大意是用 1,5,10,25,50 五种钱组成一个给定的N ,求有多少种组发。。

    dp[i]为 价值I 的个数, 那么dp[i] += dp[i-k]  (k为上面五种钱币 。。 )

    题目:

    Coin Change 

    Suppose there are 5 types of coins: 50-cent, 25-cent, 10-cent, 5-cent, and 1-cent. We want to make changes with these coins for a given amount of money.


    For example, if we have 11 cents, then we can make changes with one 10-cent coin and one 1-cent coin, two 5-cent coins and one 1-cent coin, one 5-cent coin and six 1-cent coins, or eleven 1-cent coins. So there are four ways of making changes for 11 cents with the above coins. Note that we count that there is one way of making change for zero cent.


    Write a program to find the total number of different ways of making changes for any amount of money in cents. Your program should be able to handle up to 7489 cents.

    Input 

    The input file contains any number of lines, each one consisting of a number for the amount of money in cents.

    Output 

    For each input line, output a line containing the number of different ways of making changes with the above 5 types of coins.

    Sample Input 

    11
    26
    

    Sample Output 

    4
    13
    

    代码:

     1 #include <iostream>
     2 #include <memory.h>
     3 using namespace std;
     4 
     5 int dp[10000];
     6 int M[5] = {50,25,10,5,1};
     7 int main()
     8 {
     9     int total;
    10     while(cin>>total)
    11     {
    12         if(total ==0){ cout<<"0"<<endl;continue;}
    13 
    14         dp[0]=1;
    15             for(int j = 0;j<=total;j++)
    16             {
    17                 for(int i=0;i<5;i++)
    18                 {
    19                     dp[j+M[i]] += dp[j];
    20                 }
    21         }
    22         cout<<dp[total]<<endl;
    23         memset(dp,0,sizeof(dp));
    24     }
    25 
    26 
    27     return 0;
    28 }

    代码2

     1 #include <iostream>
     2 #include <memory.h>
     3 using namespace std;
     4 
     5 int dp[10000][6];
     6 int M[5] = {1,5,10,25,50};
     7 int main()
     8 {
     9     int total;
    10 
    11 
    12         dp[0][0]=1;
    13 
    14         for(int i=1;i<=8000;i++)
    15         {
    16             for(int j=0;j<5;j++)
    17             {
    18                 for(int k=0;k<=j;k++)    //一开始这里写错了 k 必须小于j 否则会出现多加的情况
    19                 {
    20                     if( i<M[j]  )break;
    21                     dp[i][j] += dp[i-M[j]][k];
    22                 }
    23             }
    24         }
    25 
    26     while(cin>>total)
    27     {
    28         if(total ==0){ cout<<"0"<<endl;continue;}
    29 
    30         int ans=0;
    31         for(int i=0;i<5;i++)
    32         {
    33             ans+=dp[total][i];
    34         }
    35         cout<<ans<<endl;
    36 
    37     }
    38 
    39 
    40     return 0;
    41 }
  • 相关阅读:
    Spring MVC-表单(Form)标签-单选按钮(RadioButton)示例(转载实践)
    Ubuntu 16.04中VirtualBox 5.1使用U盘/USB设备的方法
    Spring MVC-表单(Form)标签-复选框集合(Checkboxes)示例(转载实践)
    Ubuntu 16.04下减小/释放/清理VirtualBox虚拟硬盘文件的大小
    关注点分离
    谈代码注释
    DelegatingFilterProxy类的作用
    GOPS 2018全球运维大会上海站 参会感悟梳理
    Java switch case
    Android 微信网址分享添加网络图片
  • 原文地址:https://www.cnblogs.com/doubleshik/p/3495667.html
Copyright © 2011-2022 走看看