母函数

母函数详解：http://blog.csdn.net/lishuhuakai/article/details/8044431

 

Square Coins

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8090    Accepted Submission(s): 5487

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
  

 

虽然说别人都说比较简单，但是我表示没有发现。。。。。。

借用别人的代码。。。。。。。。

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stdlib.h>
const int MAX=32768+10;
long long d[MAX];//#define MAX 40000
//long long d[MAX];
void solve()
{
   int i,j,n;
   while(~scanf("%d",&n),n)              
   {
    memset(d,0,sizeof(d));
    d[0]=1;
      for(i=1;i<=floor(sqrt(n));i++)
   {                                                                                 看的我是很费解啊。。。。
    for(j=i*i;j<=n;j++)
     d[j]+=d[j-i*i];
   }
    printf("%lld\n",d[n]); 
   }
}
int main(void)
{
  solve();
  return 0;
}