HDU 1028 Ignatius and the Princess III （母函数 ， DP）

Ignatius and the Princess III

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 22045 Accepted Submission(s): 15395

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
可以看博客： 链接 这里写链接内容

题目代码：

#include
#include
#include
using namespace std;
int main()
{
    int i,j,k,n,text[1010],text1[1010];
    while(scanf("%d",&n)!=EOF)
    {
        memset(text1,0,sizeof(text1));
        for(int i=0;i<=n;i++)
            text[i]=1;
        for(int i=2;i<=n;i++) //括号数
        {
            for(int j=0;j<=n;j++)//j和k是相乘时间双方的指数
            {
                for(int k=0;j+k<=n;k+=i)
                    text1[j+k]+=text[j];//数组储存的都是系数
            }
            for(int j=0;j<=n;j++)//交换 数组 将text1数组初始化
            {
                text[j]=text1[j];
                text1[j]=0;
            }

        }
         printf("%d\n",text[n]);
    }
   return 0;
}

大佬用 一维dp 写的：

/Hdu 1028
#include
#include
#include
#include
using namespace std;
long long int a[100100];
int main()
{
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        for(int i=0;i<=n;i++)
            a[i]=0;
        a[0]=1;
        for(int i=1;i<=n;i++)
        {
            for(int j=i;j<=n;j++)
            {
                a[j]+=a[j-i];
            }
        }
        printf("%lld\n",a[n]);
    }
    return 0;
}

转载于:https://www.cnblogs.com/nanfenggu/p/7900087.html