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): 9729 Accepted Submission(s): 6876


[align=left]Problem Description[/align]
"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!"

[align=left]Input[/align]
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.

[align=left]Output[/align]
For each test case, you have to output a line contains an integer P which indicate the different equations you have found.

[align=left]Sample Input[/align]

4
10
20

[align=left]Sample Output[/align]

5
42
627

[align=left]Author[/align]
Ignatius.L
母函数模板题

#include<stdio.h>
int c1[1000],c2[1000];
int main()
{
int n,i,j,k,c;
while(scanf("%d",&n)!=EOF)
{
for(i=0;i<=n;i++)
{
c1[i]=0;
c2[i]=1;
}
for(i=1;i<=n;i++)
{
for(j=0;j<=n;j++)
{
for(k=0;k+j<=n;k+=i)
{
c2[j+k]+=c1[j];
}
}
for(c=0;c<=n;c++)
{
c1[c]=c2[c];
c2[c]=0;
}
}
printf("%d\n",c1
);

}
return 0;
}

View Code