HDU1398Square Coins（母函数）

母函数介绍见另一篇随笔HDU1028Ignatius and the Princess III(母函数)

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<map>
#include<vector>
#include<set>
#include<stack>
#include<queue>
#include<algorithm>
#include<stdlib.h>
using namespace std;
#define MAX(a,b) (a > b ? a : b)
#define MIN(a,b) (a < b ? a : b)
#define MAXN 400005
#define INF 2000000007
#define mem(a) memset(a,0,sizeof(a))

int c1[301],c2[301];

int main()
{
    int n;
    while(~scanf("%d",&n) && n!= 0)
    {
        int i;
        for(i =0 ;i<=n;i++)
        {
            c1[i] = 1;
            c2[i] = 0;
        }
        for(i =2 ;i<=17; i++)
        {
            for(int j = 0;j<=n;j++)
            {
                for(int k = 0; k+j <= n; k+=i*i)
                {
                    c2[j+k]+=c1[j];
                }
            }
            for(int j = 0; j<=n;j++)
            {
                c1[j] = c2[j];
                c2[j] = 0;
            }
        }
        printf("%d
",c1[n]);
    }
    return 0;
}

另外我也写了一个其他方法，可以看看

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<map>
#include<vector>
#include<set>
#include<stack>
#include<queue>
#include<algorithm>
#include<stdlib.h>
using namespace std;
#define MAX(a,b) (a > b ? a : b)
#define MIN(a,b) (a < b ? a : b)
#define MAXN 400005
#define INF 2000000007
#define mem(a) memset(a,0,sizeof(a))

int ans(int n,int k)
{
    if(k == 1)return 1;
    int x = n/(k*k);
    int num=0;
    for(int i =0 ;i<= x; i++)
    {
        num += ans(n-i*k*k, k-1);
    }
    return num;
}

int main()
{
    int n;
    while(~scanf("%d",&n) && n != 0)
    {
        int i,key;
        for(i =1;i<=18;i++)
        {
            if(i*i > n){
                key = i-1;
                break;
            }
        }
        printf("%d
",ans(n,key));
    }
    return 0;
}