HDU--1028--Ignatius and the Princess III--全然背包

Ignatius and the Princess III

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

Total Submission(s): 11896    Accepted Submission(s): 8424

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：把一个数因式分解，点到顺序的不算。比方5=1+4和5=4+1就算是一种方式

题解：等于在式子的‘+’号后面一次加上一个数字形成新的，并且要求加上的这个数字是当前式子里面最大的，这样就不会反复

如：（去想象这个表中的数据是一行一行从左向右刷出来的“一行的定义是每一个+n算一行”，每次刷的位置的各种情况都是依据先前已经算出来了的数据得来的）

分解的那个整数 1 2 3 4 5 6

+1 1   1+1 1+1+1 1+1+1+1 1+1+1+1+1 1+1+1+1+1+1

————————————————————————————————————————

+2 2 1+2 1+1+2 1+1+1+2 1+1+1+1+2

2+2 1+2+2 1+1+2+2

2+2+2

————————————————————————————————————————

+3 3 1+3 1+1+3 1+1+1+3

2+3 1+2+3

3+3

————————————————————————————————————————

+4 4 1+4 1+1+4

2+4

————————————————————————————————————————

+5 5 1+5

————————————————————————————————————————

+6 6

————————————————————————————————————————

因式分解的方式数：1 2 3 5 7 11

————————————————————————————————————————

比方+1的时候。6就从5里面找，由于5+1=6，+2的时候6就从4里面找。+3的时候就从3里面找，这么算下去，到+6的时候，6本身也算进来，所以就等于从0里面加一个

#include <iostream>
using namespace std;
int main (void)
{
    int i,j,d[125]={1};
    for(i=1;i<121;i++)	//依次选择+n
    for(j=i;j<121;j++)	//从能+n的最小的那个数開始+n
    d[j]+=d[j-i];
    while(cin>>i)
    cout<<d[i]<<endl;
    return 0;
}