杭电1398-Square Coins
来源:互联网 发布:语音识别用到的算法 编辑:程序博客网 时间:2024/05/17 18:25
Square Coins
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 6624 Accepted Submission(s): 4473
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.
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
210300
Sample Output
1427AC代码+详细解释:#include<iostream>//母函数做法,排列组合加法对应指数幂乘法#include<cstdio>#include<cstring>const int MAX=301;int c1[MAX];//c1[2]=1代表价值2的支付金额只有一种方式来付即文中two 1-credit coins仅此一种,同理c1[n]=m代表价值为n的支付金额可以有m种付法int c2[MAX];//中间变量int n;using namespace std;int main(){ int i,j; while(cin>>n,n) { for(i=0;i<=n;i++) { c1[i]=1; c2[i]=0; } for(i=2;i<=17;++i)//一共有多少种硬币 { for(j=0;j<=n;j++)//这里j是指数即与之相乘的指数 { for(int k=0;k+j<=n;k+=i*i)//k+j代表相乘后指数相加 { c2[j+k]+=c1[j]; } } for(j=0;j<=n;j++) { c1[j]=c2[j];//把过渡量再赋值给c1 c2[j]=0;//再初始化为0 } } cout<<c1[n]<<endl; } return 0;}
- 杭电1398-Square Coins
- 杭电1398-Square Coins
- 杭电 HDU ACM 1398 Square Coins
- 杭电hdu 1398 Square Coins 母函数
- 杭电ACM 1398 Square Coins(母函数)
- 杭电1398 Square Coins(母函数)
- Square Coins(杭电1398)(母函数)
- 杭电1398 Square Coins 简单母函数
- 杭电ACM hdu 1398 Square Coins 解题报告(母函数)
- 【杭电oj】1398 - Square Coins(母函数打表)
- 杭电ACM1398——Square Coins~~母函数
- hdu 1398Square Coins
- 1398 Square Coins
- hdu 1398 Square Coins
- Hdu 1398 - Square Coins
- hdu 1398 Square Coins
- hdu 1398 Square Coins
- hdu 1398 Square Coins
- 一个可无限伸缩且无ABA问题的无锁队列
- 大连实训之基于MFC对数据库的基本操作(功能实现)
- KMP算法详解
- 一天Android App[简单钢琴]开发 [李园7舍_404]
- 关于android WebViewClient的方法解释
- 杭电1398-Square Coins
- Android4.0 隐藏虚拟按键 实现全屏
- UNIX网络编程——ICMP报文分析:端口不可达
- hdu 2795 Billboard 线段树基础题
- hibernate ——get save
- hdu4607
- 西山居决赛
- Vmware的三种设置网络设置Bridge, …
- ubuntu10.04 中安装 Vmware Tools