【杭电oj】1028 - Ignatius and the Princess III（母函数打表）

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Ignatius and the Princess III

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 18586    Accepted Submission(s): 13020


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

代码如下：

#include <cstdio>
#include <cstring>
#define MAX 120
int main()
{
int c[MAX+10];
int t[MAX+10];
memset (c,0,sizeof (c));
memset (t,0,sizeof (t));
for (int i = 0 ; i <= MAX ; i++)
c[i] = 1;
for (int i = 2 ; i <= MAX ; i++)
{
for (int j = 0 ; j <= MAX ; j++)
for (int k = 0 ; k+j <= MAX ; k += i)
t[j+k] += c[j];
for (int j = 0 ; j <= MAX ; j++)
{
c[j] = t[j];
t[j] = 0;
}
}
int n;
while (~scanf ("%d",&n))
printf ("%d\n",c
);
return 0;
}