HDU1398--Square Coins（母函数）

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

 

母函数的模版题目，改变的只是模拟多项式乘法循环时的步进

代码：

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>

using namespace std;

const int MAX = 300;

int ele[18];
int ans[MAX+2],tmp[MAX+2];

void work(){
    int i,j,k;

    memset(ans,0,sizeof(ans));
    memset(tmp,0,sizeof(tmp));

    for(i=0;i<=MAX;++i){
        ans[i]=1;
    }

    for(i=1;i<17;++i){
        for(j=0;j<=MAX;++j){
            for(k=0;k<=MAX && j+k*ele[i]<=MAX;++k){
                tmp[j+k*ele[i]] += ans[j];
            }
        }
        for(j=0;j<=MAX;++j){
            ans[j] = tmp[j];
            tmp[j] = 0;
        }
    }
}

void init(){
    int i;

    for(i=1;i<=17;++i){
        ele[i-1] = i*i;
    }

    work();
}



int main(){
    //freopen("in.txt","r",stdin);
    //(author : CSDN iaccepted)
    init();

    int n;
    while(scanf("%d",&n)!=EOF){
        if(n==0)break;
        printf("%d
",ans[n]);
    }
    return 0;
}