HDU 1028 Ignatius and the Princess III（母函数模板）

Ignatius and the Princess III

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

Total Submission(s): 17839 Accepted Submission(s): 12508



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

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母函数初学者，还是看大神的总结吧http://blog.csdn.net/vsooda/article/details/7975485，下面是个模板题。。

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int main()
{
int n,i,j,k;
int c1[140];//保存每个数字可以组合的数目
int c2[140];//中间量，保存每一次的情况
while(scanf("%d",&n)!=EOF)
{
for(i=0;i<=n;i++)//初始化
{
c1[i]=1;
c2[i]=0;
}
for(i=2;i<=n;i++)
{
for(j=0;j<=n;j++)
{
for(k=0;k+j<=n;k+=i)

c2[k+j]+=c1[j];

}
for(j=0;j<=n;j++)
{
c1[j]=c2[j];
c2[j]=0;
}
}
printf("%d\n",c1
);
}
return 0;
}