zoukankan      html  css  js  c++  java
  • HDU 1398 Square Coins

    Square Coins

    Problem Description
    People in Silverland use square coins. Not only they have square shapes but also their values are square numbers. Coins with values of all square numbers up to 289 (=17^2), i.e., 1-credit coins, 4-credit coins, 9-credit coins, ..., and 289-credit coins, are available in Silverland. 
    There are four combinations of coins to pay ten credits: 

    ten 1-credit coins,
    one 4-credit coin and six 1-credit coins,
    two 4-credit coins and two 1-credit coins, and
    one 9-credit coin and one 1-credit coin. 

    Your mission is to count the number of ways to pay a given amount using coins of Silverland.
     
    Input
    The input consists of lines each containing an integer meaning an amount to be paid, followed by a line containing a zero. You may assume that all the amounts are positive and less than 300.
     
    Output
    For each of the given amount, one line containing a single integer representing the number of combinations of coins should be output. No other characters should appear in the output. 
     
    Sample Input
    2 10 30 0
     
    Sample Output
    1 4 27
    :用母函数求解
    #include<iostream>
    using namespace std;
    #include<math.h>
    int main()
    {
        int n;
        int i,j,k,p=0;
        int c[20],a[305],b[305];
        for(i=1;i<=300;i++)
        {
            if(pow((int)(sqrt(i*1.0)),2)==i) //sqrt(i*1.0)
            {
             c[p++]=i;
            }
        }
    
        while(scanf("%d",&n)!=EOF&&n)
        {
            for(i=0;i<=n;i++)
                {
                    a[i]=1;  //1
                    b[i]=0;
                }
            for(i=1;i<p;i++)
            {
                for(j=0;j<=n;j++)
                 for(k=0;k+j<=n;k+=c[i])
                 {
                     b[k+j]+=a[j];
                 }
                 for(j=0;j<=n;j++)
                 {
                     a[j]=b[j];//2
                     b[j]=0;
                 }
            }
            printf("%d\n",a[n]);
        }
        
        return 0;
    }
     
  • 相关阅读:
    使用vs2010编译 Python SIP PyQt4
    谷歌编程指南
    【转】微策略面经相关资料
    KMP 算法
    C++ 拷贝构造函数
    虚继承 虚表 定义一个不能被继承的类
    cache的工作原理
    背包问题
    【转】C/C++ 内存对齐
    【转】 Linux/Unix 进程间通信的各种方式及其比较
  • 原文地址:https://www.cnblogs.com/hsqdboke/p/2455431.html
Copyright © 2011-2022 走看看