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[],c2[];

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

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

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