hdu1398Square Coins

Square Coins

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11567    Accepted Submission(s): 7921

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

210300

 

Sample Output

1427
(1+x+x^2+x^3+...+x^n)(1+x^4+x^8+...)(1+x^9+x^18+..)...(1+x^sqrt(n))最多可以乘到sqrt(n)项，所以对于i展开时从2到sqrt(n)即可，但是犯了一个错误，sqrt（）返回的是一个double，不是整数，应该写成i*i<=n
#include<stdio.h>#include<math.h>int main(){int c1[310],c2[310];int i,j,k,n;while(scanf("%d",&n),n){for(i=0;i<=n;i++){c1[i]=1;c2[i]=0;}for(i=2;i*i<=n;i++){for(j=0;j<=n;j++)for(k=0;k+j<=n;k+=i*i)c2[j+k]+=c1[j];for(j=0;j<=n;j++){c1[j]=c2[j];c2[j]=0;}}printf("%d\n",c1[n]);}return 0;}