Square Coins

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12929    Accepted Submission(s): 8885

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

Source

Asia 1999, Kyoto (Japan)

Recommend

Ignatius.L   |   We have carefully selected several similar problems for you:  1171 1085 2152 2082 1709

母函数代码：

#include<iostream>
using namespace std;
#define M 1000
#define MAXN 310      //MAXN为最大有多少项相乘
int a[M],b[M];//a[M]中存最终项系数;b[M]中存取中间变量;
int main()
{
    int m,n;
    int i,j,k;
    m=MAXN;
    while(scanf("%d",&n)!=EOF)
    {
        if(n==0)break;
        //可以加一个跳出循环的条件
        //n为所求的指数的值: ";
        //因为只求指数为n的系数的值:所以循环只到n就结束
        for(i=0;i<=n;i++)//初始化第一个式子:(1+X^2+X^3+...) 所以将其系数分别存到a[n]
        {
            a[i]=1;
            b[i]=0;
        }
        for(i=2;i<=17;i++)//从第2项式子一直到第n项式子与原来累乘项的和继续相乘
        {
            for(j=0;j<=n;j++)//从所累乘得到的式子中指数为0遍历到指数为n 分别与第i个多项式的每一项相乘
                 for(k=0;k+j<=n;k+=i*i)//第i个多项式的指数从0开始,后面的每项指数依次比前面的多i,比如当i=3时,第3项的表达式为(1+x^3+x^6+x^9+……),直到所得指数的值i+j>=n退出
                 {
                     b[j+k]+=a[j];//比如前面指数为1,系数为3,即a[1]=3 的一项和下一个表达式的指数为k=3的相乘,则得到的式子的系数为,b[j+k]=b[4]+=a[1],又a[1]=3,所以指数为4的系数为b[4]=3;
                 }

             for(j=0;j<=n;j++)//  然后将中间变量b数组中的值依次赋给数组a,然后将数组b清零,继续接收乘以下一个表达式所得的值
              {
                  a[j]=b[j];
                  b[j]=0;
              }
        }
        printf("%d\n",a[n]);  // 指数为n的项的系数为:
    }

    return 0;
}

DP代码：

#include<iostream>
using namespace std;
#define M 1000
int dp[300]={1};
int main()
{
    int n;
     for(int i=1;i<=17;i++)
        for(int j=i*i;j<=300;j++)
          dp[j]+=dp[j-i*i];
        while(scanf("%d",&n)!=EOF)
        {
          if(n==0)break;
           printf("%d\n",dp[n]);
        }
    return 0;
}