hdu 1028 Ignatius and the Princess III（母函数）

Ignatius and the Princess III

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

Total Submission(s): 9848 Accepted Submission(s): 6962



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

题意：拆分整数

题解：构造母函数G（x）=（1+x+……）（1+x^2+……）……即可

#include<stdio.h>
int dp[128],temp[128];
int main()
{
int i,j,k,n;

while(scanf("%d",&n)>0)
{
for(i=0;i<=n;i++) dp[i]=1,temp[i]=0;
for(i=2;i<=n;i++)
{
for(j=0;j<=n;j++)
{
for(k=0;k+j<=n;k+=i)
temp[j+k]+=dp[j];
}
for(j=0;j<=n;j++) dp[j]=temp[j],temp[j]=0;
}
printf("%d\n",dp
);
}

return 0;
}