hdu1028(dp 或母函数)

最新推荐文章于 2019-03-12 18:00:56 发布

原创最新推荐文章于 2019-03-12 18:00:56 发布 · 372 阅读

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本文探讨了一个典型的计数问题，即求解整数N的不同拆分方式的数量，并通过动态规划算法提供了解决方案。文章提供了完整的代码实现，展示了如何通过逐步累加的方式避免重复计算。

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Ignatius and the Princess III

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 24446 Accepted Submission(s): 16980

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
 

Author

Ignatius.L

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题意：给你一个数n,问n有多少总凑法让这些数的和等于n

例如 3=1+1+1=1+2=3 有三种

题解：

1.母函数模板（第二类问题）

2.dp

显然凑数的时候大的数方法会覆盖小的数，那么凑大的数就可以往前找

例如凑5每次加一的话就往4加，加二就往3加

#include <iostream>
using namespace std;
int main (void)
{
int i,j,d[125]={1};//d[0]=1，其他自动初始化为0

for(i=1;i<121;i++) //从小到大加不会重复，从1往后加
for(j=i;j<121;j++) //从能+i的最小的那个数开始+n
d[j]+=d[j-i];//加到自身就是加d[0] （ =1 ）
while(cin>>i)
cout<<d[i]<<endl;
return 0;
}