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

 1 #include<stdio.h>
 2 #include<string.h>
 3 int num_[121] , num[121] ;
 4 int main ()
 5 {
 6     int n ;
 7     while ( scanf ( "%d" , &n ) != EOF )
 8     {
 9           for ( int i = 0 ; i <= n ; ++ i )
10           {
11               num_[i] = 0 ;
12               num[i] = 1 ;
13           } 
14           for ( int i = 2 ; i <= n ; ++ i ) 
15           {
16               for ( int j = 0 ; j <= n ; ++ j )
17                   for ( int k = 0 ; k + j <= n ; k += i )
18                   {
19                       num_[j+k] += num[j] ;
20                   }
21               for ( int j = 0 ; j <= n ;  ++ j )
22               {
23                   num[j] = num_[j] ;
24                   num_[j] = 0 ;
25               }
26           }
27           printf ( "%d\n" , num[n] ) ;
28     }
29     return 0 ; 
30 }

　该题目还可以用完全背包写：

 1 #include<stdio.h>
 2 #include<string.h>
 3 int dp[121] , n ;
 4 void get_dp ()
 5 {
 6      memset( dp , 0 , sizeof (dp) ) ;
 7      dp[0] = 1 ;
 8      for ( int i = 1 ; i <= 120 ; ++ i )
 9          for ( int j = i ; j <= 120 ; ++ j )
10              dp[j] += dp[j - i] ; 
11 } 
12 int main ()
13 {
14     get_dp() ;
15     while ( scanf ( "%d" , &n ) != EOF )
16     {
17           printf ( "%d\n" , dp[n] ) ;
18     }
19     return 0 ;
20 }