(Simple parent function template) Hdu 1028 Ignatius and the Princess III

Ignatius and the Princess IIITime limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others) Total Submission (s): 14405 Accepted Submission (s): 10142

Problem Description "Well, it seems the first problem are 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! "

Inputthe 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.

Outputfor each of the test case and you had to output a line contains an integer P which indicate the different equations you had Found.

Sample Input

41020

Sample Output

542627

Authorignatius.l
Examine the knowledge point; Parent function template

Female function template--integer split--infinite number of coins//facts: The Memset function cannot be assigned to an array other than 0,-1 1 #include <stdio.h> #include <string.h>int c1[ 125],c2[125];int Main () {int N;int i,j,k;while (~scanf ("%d", &n)) {for (i=0;i<=n;++i) {c1[i]=1;c2[i]=0;} for (i=2;i<=n;++i)//i expression {for (J=0;J<=N;++J)//The result of the multiplication of the preceding I expression {for (k=0;k+j<=n;k+=i)//index {C2[K+J]+=C1[J] ;}} for (J=0;J<=N;++J) c1[j]=c2[j],c2[j]=0;} printf ("%d\n", C1[n]);} return 0;}

(simple parent function template) Hdu 1028 Ignatius and the Princess III