HDU 1028 Ignatius and the Princess III

Ignatius and the Princess III

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4636 Accepted Submission(s): 3246

Problem Description

"Well, it seems the first problem is too easy. I will let you know how foolish you are later." feng5166 says.

"The second problem is, given an positive integer N, we define an equation like this:
N=a[1]+a[2]+a[3]+...+a[m];
a[i]>0,1<=m<=N;
My question is how many different equations you can find for a given N.
For example, assume N is 4, we can find:
4 = 4;
4 = 3 + 1;
4 = 2 + 2;
4 = 2 + 1 + 1;
4 = 1 + 1 + 1 + 1;
so the result is 5 when N is 4. Note that "4 = 3 + 1" and "4 = 1 + 3" is the same in this problem. Now, you do it!"

Input

The input contains several test cases. Each test case contains a positive integer N(1<=N<=120) which is mentioned above. The input is terminated by the end of file.

Output

For each test case, you have to output a line contains an integer P which indicate the different equations you have found.

Sample Input

4 10 20

Sample Output

5 42 627

Author

Ignatius.L

 

母函数 模板题 详情

#include<stdio.h>
#include<string.h>
int num_[121] , num[121] ;
int main ()
{
    int n ;
    while ( scanf ( "%d" , &n ) != EOF )
    {
          for ( int i = 0 ; i <= n ; ++ i )
          {
              num_[i] = 0 ;
              num[i] = 1 ;
          } 
          for ( int i = 2 ; i <= n ; ++ i ) 
          {
              for ( int j = 0 ; j <= n ; ++ j )
                  for ( int k = 0 ; k + j <= n ; k += i )
                  {
                      num_[j+k] += num[j] ;
                  }
              for ( int j = 0 ; j <= n ;  ++ j )
              {
                  num[j] = num_[j] ;
                  num_[j] = 0 ;
              }
          }
          printf ( "%d\n" , num[n] ) ;
    }
    return 0 ; 
}

　该题目还可以用完全背包写：

#include<stdio.h>
#include<string.h>
int dp[121] , n ;
void get_dp ()
{
     memset( dp , 0 , sizeof (dp) ) ;
     dp[0] = 1 ;
     for ( int i = 1 ; i <= 120 ; ++ i )
         for ( int j = i ; j <= 120 ; ++ j )
             dp[j] += dp[j - i] ; 
} 
int main ()
{
    get_dp() ;
    while ( scanf ( "%d" , &n ) != EOF )
    {
          printf ( "%d\n" , dp[n] ) ;
    }
    return 0 ;
}