HDU 1028(母函数)-LMLPHP
Online JudgeOnline ExerciseOnline TeachingOnline ContestsExercise Author
F.A.Q
Hand In Hand
Online Acmers
Forum |Discuss
Statistical Charts
Problem Archive
Realtime Judge Status
Authors Ranklist
 
     C/C++/Java Exams
ACM Steps
Go to Job
Contest LiveCast
ICPC@China
Best Coder 
VIP | STD Contests
Virtual Contests
  DIY |Web-DIY 
Recent Contests
 

Ignatius and the Princess III

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 19181    Accepted Submission(s): 13477

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

题意:给你一个正整数n,将该正整数拆分成若干整数的和,问你共有多少不同的拆分方法?

#include<bits/stdc++.h>
using namespace std;
int c1[],c2[];//c1用于记录各项前面的系数,c2用于记录中间值
int main()
{
int n;
while(~scanf("%d",&n))
{
//共有n个括号,先处理第一个括号,第一个括号里每一项的系数都是1;
for(int i=;i<=n;i++)
{
c1[i]=;
c2[i]=;
}
//因为共有n个括号,i表示正在处理第i个括号
for(int i=;i<=n;i++)
{
//j表示正在处理第i个括号里的第就项,
for(int j=;j<=(n);j++)
{
for(int k=;k+j<=n;k+=i)
{
c2[j+k]+=c1[j];
}
}
for(int j=;j<=(n);j++)
{
c1[j]=c2[j];
c2[j]=;
}
}
printf("%d\n",c1[n]);
}
return ;
}
05-11 16:26