hdu 1398 Square Coins

最新推荐文章于 2019-07-18 02:03:39 发布

转载最新推荐文章于 2019-07-18 02:03:39 发布 · 54 阅读

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原文链接：http://www.cnblogs.com/duan-to-success/p/3474916.html

本文介绍了一个使用平方数硬币支付给定金额的问题，并提供了一段C语言代码实现。该问题探讨了不同面额平方数硬币组合的方式数量，通过多项式函数的方法进行计算。

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.

此题和1085题没什么区别，思路是一样的。都是母函数基本的应用。

#include<stdio.h>
#include<string.h>
int c1[301],c2[301];
int main()
{
    int n,i,j,k;
    int a[17]={1,4,9,16,25,36,49,64,81,100,121,144,13*13,14*14,15*15,16*16,17*17};
    while(scanf("%d",&n)!=EOF&&n!=0)
    {
        memset(c1,0,sizeof(c1));
        memset(c2,0,sizeof(c2));
        for(i=0;i<=n;i++)
            c1[i]=1;
        for(i=2;i<=17;i++)//i表示第i个表达式
        {
            for(j=0;j<=n;j++)//j指的是前面i-1个表达式相乘后得到的表达式每一项的指数
                for(k=0;k+j<=n;k+=a[i-1])//k表示第i个表达式中每一项的指数
                    c2[k+j]+=c1[j];
            for(j=0;j<=n;j++)
            {
                c1[j]=c2[j];
                c2[j]=0;
            }
        }
        printf("%d\n",c1[n]);
    }
    return 0;
}

转载于:https://www.cnblogs.com/duan-to-success/p/3474916.html