hdu 1028 Ignatius and the Princess III

最新推荐文章于 2020-03-06 19:09:17 发布

处变不惊

最新推荐文章于 2020-03-06 19:09:17 发布

阅读量384

点赞数

CC 4.0 BY-SA版权

分类专栏： ACM试题

本文链接：https://blog.youkuaiyun.com/zyq_19960204/article/details/50820376

ACM试题专栏收录该内容

104 篇文章

订阅专栏

本文探讨了给定正整数N的整数划分问题，即寻找所有可能的加法组合使得这些组合的和等于N。文章提供了两种不同的编程实现方法：一种基于母函数的方法，另一种则是使用动态规划技术。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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
(母函数)

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<algorithm>
using namespace std;
int main()
{
    int a[500],b[500],i,j,k,n;
    while (~scanf("%d",&n))
    {
        for (i=0;i<=n;i++)
          {
              a[i]=1;       //赋初值
              b[i]=0;
          }
          for (i=2;i<=n;i++)      //根据情况进行变换(i*i<=n)
           {
               for (j=0;j<=n;j++)
                 for (k=0;k+j<=n;k+=i)
                    b[k+j]+=a[j];       //这两个for循环将i的一种情况进行了统计
               for (j=0;j<=n;j++)
               {
                   a[j]=b[j];       //进行传值，并将b数组清零
                   b[j]=0;
               }
           }
        printf("%d\n",a[n]);
    }
    return 0;
}

(dp动态规划)

#include<stdio.h>
__int64 dp[125];
void fun(int n)
{
   int i,j;
   dp[0]=1;
  for (i=1;i<=n;i++)
   for (j=i;j<=n;j++)
    dp[j]+=dp[j-i];<span style="white-space:pre">	</span>//意思是将比它小的值的和加起来赋给比它大的值
}
int main()
{
    int m;
    fun(120);
   while (~scanf("%d",&m))
    printf("%I64d\n",dp[m]);
    return 0;
}