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!"
"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 }