HDU 1028 Ignatius and the Princess III

Description

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

"The second problem is, given an positive integer N, we define a equation like this:
N=A[1]+A[2]+A[3]+...+A[M];
a[i]>0,1<=m<=n;
My question is what 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;
The result is 5 while N is 4. Note that "4 = 3 + 1" and "4 = 1 + 3" are the same in this problem. Now, do it! "

Input

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

Output

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

Sample Input

41020

Sample Output

542627 code is ZXP that moved the completely unknown Li = =

#include <cstdio>#include<cstring>#include<iostream>#include<algorithm>using namespaceStd;__int64 a[ Max];intMain () {inti,j; a[0]=1;//a[1]=1;a[2]=2;a[3]=3;a[4]=5;a[5]=7;     for(i=1; i<= -; i++)    {         for(j=0; i+j<= -; j + +) {A[i+j]+=A[j]; }    }    intN;  while(SCANF ("%d", &n)! =EOF) {printf ("%i64d\n", A[n]); }    return 0;}

View Code

HDU 1028 Ignatius and the Princess III