https://vjudge.net/contest/67836#problem/G

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

时间复杂度:

题解:动态规划

代码:

#include <bits/stdc++.h>
using namespace std; int coin[20] = {1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289};
int a[333][333];
int money; int main() {
while(~scanf("%d", &money)) {
memset(a, 0, sizeof(a));
if(money == 0)
break;
a[0][0] = 1;
for(int i = 0; i < 17; i ++) {
for(int j = coin[i]; j <= money; j ++) {
for(int k = 1; k < 300; k ++) {
if(j >= coin[i])
a[k][j] += a[k - 1][j - coin[i]];
}
}
} int sum = 0;
for(int i = 0; i < 300; i ++)
sum += a[i][money]; printf("%d\n", sum);
}
return 0;
}

  

ZOJ 1666 G-Square Coins的更多相关文章

  1. hdu 1398 Square Coins (母函数)

    Square Coins Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Tota ...

  2. HDOJ 1398 Square Coins 母函数

    Square Coins Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Tota ...

  3. 题解报告:hdu 1398 Square Coins(母函数或dp)

    Problem Description People in Silverland use square coins. Not only they have square shapes but also ...

  4. hdu 1398 Square Coins(简单dp)

    Square Coins Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Pro ...

  5. Square Coins[HDU1398]

    Square Coins Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Tota ...

  6. Square Coins(母函数)

    Square Coins 点我 Problem Description People in Silverland use square coins. Not only they have square ...

  7. HDU1398 Square Coins(生成函数)

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission( ...

  8. HDU 1398 Square Coins 整数拆分变形 母函数

    欢迎参加——BestCoder周年纪念赛(高质量题目+多重奖励) Square Coins Time Limit: 2000/1000 MS (Java/Others)    Memory Limit ...

  9. HDU1398 Square Coins

    Description People in Silverland use square coins. Not only they have square shapes but also their v ...

随机推荐

  1. 全文检索引擎 sphinx-coreseek中文索引

    Sphinx是一个基于SQL的全文检索引擎,可以结合MySQL,PostgreSQL做全文搜索,它可以提供比数据库本身更专业的搜索功能,使得应用程序更容易实现专业化的全文检索. Sphinx特别为一些 ...

  2. LinkedList的源码分析(基于jdk1.8)

    1.初始化 public LinkedList() { } 并未开辟任何类似于数组一样的存储空间,那么链表是如何存储元素的呢? 2.Node类型 存储到链表中的元素会被封装为一个Node类型的结点.并 ...

  3. QWT编译与配置-Windows/Linux环境

    QWT编译与配置-Windows/Linux环境 QWT和FFTW两种开源组件是常用的工程软件支持组件,QWT可以提供丰富的绘图组件功能,FFTW是优秀数字波形分析软件.本文使用基于LGPL版权协议的 ...

  4. Redis系列化方式有哪些?哪个系列化性能最好?

    Redis系列化方式有JDK系列化.JSON系列化.XML系列化等多种.我专门测试过,在我的笔记本电脑上保存5万条User对象到Redis,JDK系列化方式平均要15秒,JSON系列化方式只要13秒多 ...

  5. 统一建模语言——UML

    一.UML概述 Unified Modeling Language (UML)又称统一建模语言或标准建模语言,是始于1997年一个OMG标准,它是一个支持模型化和软件系统开发的图形化语言,为软件开发的 ...

  6. 天津Uber优步司机奖励政策(12月14日到12月20日)

    滴快车单单2.5倍,注册地址:http://www.udache.com/ 如何注册Uber司机(全国版最新最详细注册流程)/月入2万/不用抢单:http://www.cnblogs.com/mfry ...

  7. P1113 杂务

    P1113 杂务 题目描述 John的农场在给奶牛挤奶前有很多杂务要完成,每一项杂务都需要一定的时间来完成它.比如:他们要将奶牛集合起来,将他们赶进牛棚,为奶牛清洗乳房以及一些其它工作.尽早将所有杂务 ...

  8. C#函数库

    //读取文本文件并返回内容不同的那一行         public static String different(String sOldFile, String sNewFile)         ...

  9. textview的阴影线

    android:shadowColor="#000000" android:shadowDx="1" android:shadowDy="1" ...

  10. 「日常训练」Queue(Codeforces Round 303 Div.2 D)

    简单到让人不敢相信是D题,但是还是疏忽了一点. 题意与分析 (Codeforces 545D) 题意:n人排队,当一个人排队的时间超过他需要服务的时间就会厌烦,现在要求一个最优排列使得厌烦的人最少. ...