Square Coins（母函数）

Square Coins

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9926    Accepted Submission(s): 6806

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

题意：火星上的货币有1,4,9,16,25.....2^17这17中面值的硬币，问任意给定一个不大于300的正整数面额，用这些硬币来组成此面额总共有多少种组合种数。

题解：钱可以取多次，钱的面值是x^2，最大面值是17^2=289；母函数；

代码：

 

#include<stdio.h>
#include<math.h>
const int MAXN=310;
int main(){
    int N,a[MAXN],b[MAXN];
    while(~scanf("%d",&N),N){
        for(int i=0;i<=N;i++)
            a[i]=1,b[i]=0;
        for(int i=2;i<=17;i++){//最大硬币值是289 
            for(int j=0;j<=N;j++)
            for(int k=0;k+j<=N;k+=i*i)
                b[j+k]+=a[j];
            for(int j=0;j<=N;j++)
                a[j]=b[j],b[j]=0;
        }
            printf("%d
",a[N]);
    }
    return 0;
} 

extern "C++"{
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
using namespace std;
const int INF = 0x3f3f3f3f;
#define mem(x,y) memset(x,y,sizeof(x))
typedef long long LL;
typedef unsigned long long ULL;

void SI(int &x){scanf("%d",&x);}
void SI(double &x){scanf("%lf",&x);}
void SI(LL &x){scanf("%lld",&x);}
void SI(char *x){scanf("%s",x);}

}
const int MAXN = 350;
int a[MAXN],b[MAXN];
int x[30];
int main(){
	int N;
	while(scanf("%d",&N),N){
		for(int i = 0;i <= N;i++)a[i] = 1,b[i] = 0;
		for(int i = 2;i <= 25;i++)x[i] = i * i;
		for(int i = 2;x[i] <= N;i++){
			for(int j = 0;j <= N;j++){
				for(int k = 0;j + k <= N;k += x[i]){
					b[j + k] += a[j];
				}
			}
			for(int j = 0;j <= N;j++)a[j] = b[j],b[j] = 0;
		}
		printf("%d
",a[N]);
	}
	return 0;
}