acm之母函数题目1

最新推荐文章于 2018-09-30 11:19:29 发布

Jancy1072

最新推荐文章于 2018-09-30 11:19:29 发布

阅读量541

点赞数

CC 4.0 BY-SA版权

分类专栏： acm 文章标签： acm 母函数

本文链接：https://blog.youkuaiyun.com/shelly1072/article/details/51682972

acm 专栏收录该内容

36 篇文章

订阅专栏

本文介绍了一个使用动态规划解决的问题——如何用平方数价值的硬币组合支付特定金额的方法。问题来源于一个虚构的地方Silverland，这里的硬币价值为1到289之间的所有平方数。文章提供了一段C语言代码实现，通过递进地计算每种面额加入后的支付方式数量，最终得出任意给定金额的支付组合总数。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

代码：

#include<stdio.h>
int main(){
    int n;
    int c1[301];
    int c2[301];
    int i,j,k;
    for(i=0;i<=300;i++){
        c1[i]=1;
        c2[i]=0;
    }
    for(i=2;i<=17;i++){ 
        for(j=0;j<=300;j++){
            for(k=0;k+j<=300;k=k+i*i){ 
                c2[j+k]=c2[j+k]+c1[j];
            }               
        }
        for(j=0;j<=300;j++){
            c1[j]=c2[j];
            c2[j]=0;    
        }
    }
    scanf("%d",&n);
    while(n!=0){
        printf("%d\n",c1[n]);
        scanf("%d",&n);
    }
    return 0;
}