Ignatius and the Princess III
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 16742 Accepted Submission(s): 11783
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
//
// main.cpp
// 数的拆分(母函数的应用)
//
// Created by 张嘉韬 on 16/1/26.
// Copyright © 2016年 张嘉韬. All rights reserved.
//
#include <iostream>
using namespace std;
int main(int argc, const char * argv[]) {
//freopen("")
int n,c1[200],c2[200];
while(scanf("%d",&n)==1)
{
for(int i=0;i<=n;i++)
{
c1[i]=1;
c2[i]=0;
}
for(int i=2;i<=n;i++)
{
for(int j=0;j<=n;j++)
{
for(int k=0;j+k<=n;k=k+i)
{
c2[k+j]+=c1[j];
}
}
for(int j=0;j<=n;j++)
{
c1[j]=c2[j];
c2[j]=0;
}
}
cout<<c1[n]<<endl;
}
return 0;
}
母函数解析http://blog.csdn.net/vsooda/article/details/7975485