首页 > 代码库 > HDU--1028--Ignatius and the Princess III--完全背包

HDU--1028--Ignatius and the Princess III--完全背包

2024-07-03 07:47:05   228人阅读

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 12 3 4 5 6
+1 1  1+1 1+1+1 1+1+1+11+1+1+1+11+1+1+1+1+1
————————————————————————————————————————
+2 21+2 1+1+2 1+1+1+21+1+1+1+2
2+21+2+2 1+1+2+2
2+2+2

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

+3 31+31+1+31+1+1+3

2+31+2+3

3+3

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

+4 41+4 1+1+4

2+4

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

+5 51+5

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

+6 6

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

因式分解的方式数：1 2 3 57 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;
}


联系
我们