hdu 1028

母函数水过：

#include <stdio.h>
#include <string.h>
long solve(int n)
{
     int i ,j ,k;
     long c[121] ,a[121];
     memset(a ,1 ,sizeof(a));
     for(i = 0 ;i <= n ;i++)
           a[i] = 1;
     for(i = 2 ;i <= n ;i++)  //控制多项式的个数
     {
           memset(c ,0 ,sizeof(c));
           for(j = 0 ;j <= n ;j += i)
           {
                for(k = 0 ;k + j <= n ;k++)
                    c[j + k] += a[k];
           }
           memcpy(a ,c ,sizeof(c));
     }
     return a[n];
}
int main()
{
     int n;
     while(scanf("%d",&n) != EOF)
            printf("%ld\n",solve(n));
     return 0;
}

 

递归：

#include <stdio.h>
#include <string.h>
int f[121][121];
int solve(int n ,int k)
{
     if(k == 0) return f[n][k];
     else
     {
         if(k >= n)
         {
               if(f[n][k] == 0)
               {
                     if(f[n][n - 1] == 0) f[n][n - 1] = solve(n ,n - 1);
                     f[n][k] = f[n][n - 1] + 1;
               }
               return f[n][k];
         }
         else
         {
               if(f[n][k] == 0)
               {
                     if(f[n][k - 1] == 0) f[n][k - 1] = solve(n ,k - 1);
                     if(f[n - k][k] == 0) f[n - k][k] = solve(n - k ,k);
                     f[n][k] = f[n][k - 1] + f[n - k][k];
               }
               return f[n][k];
         }
     }
}
int main()
{
     int n;
     while(scanf("%d",&n) != EOF)
     {
          memset(f ,0 ,sizeof(f));
          printf("%d\n",solve(n ,n));
     }
     return 0;
}