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，比如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，而且要求加上的这个数字是当前式子里面最大的，这样就不会重复
如：（去想象这个表中的数据是一行一行从左向右刷出来的“一行的定义是每个+n算一行”，每次刷的位置的各种情况都是根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
————————————————————————————————————————
比如+1的时候，6就从5里面找，因为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的最小的那个数开始+n
d[j]+=d[j-i];
while(cin>>i)
cout<<d[i]<<endl;
return 0;
}
HDU--1028--Ignatius and the Princess III--完全背包