hdu1398Square Coins（母函数）

Square Coins

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10621    Accepted Submission(s): 7257

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
题意：给你个硬币让你让小于它的所有的平方根。 例9由1 4 9来兑换，10由1 4 9来兑换。问你有多少兑换方法。
#include <iostream>using namespace std;#include<stdlib.h>#include<stdio.h>#include <cstring>#include<string>#include<cmath>#include<algorithm>int c1[10000],c2[10000];int main(){    int T;    while(~scanf("%d",&T))    {        if(T==0)            return 0;        int a[10000];        int i,sum,j,k,w=0,n;        for(i=1;i<=T;i++)//算出是由哪些数来拼成的。        {            if(i*i<=T)            {                a[i-1]=i*i;                n=i;            }            else                break;        }        for(i=0;i<=T;i++)//初始化第一个括号。            c1[i]=1;         memset(c2,0,sizeof(c2));        for(i=1;i<n;i++)//控制种类的。        {            for(j=0;j<=T;j++)//控制第一个括号指数到到达的最大值。            {                for(k=0;k+j<=T;k+=a[i])                {                    c2[j+k]+=c1[j];//合并两个括号                }            }            for(j=0;j<=T;j++)            {                 c1[j]=c2[j];                 c2[j]=0;            }        }        printf("%d\n",c1[T]);    }    return 0;}