HDU1028：Ignatius and the Princess III（dp入门 & 母函数）

Ignatius and the Princess III

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 19924    Accepted Submission(s): 13936

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

41020

 

Sample Output

542627

 

Author

Ignatius.L

将一个正整数分成若干个数相加，求分解的方案数。

法一：动态规划：易得dp[i] = dp[i] + dp[i-a[k]], （k<=i）。

# include <stdio.h>int main(){    int n, i, j, dp[125]={1};    for(i=1; i<=120; ++i)        for(j=i; j<=120; ++j)            dp[j] += dp[j-i];    while(~scanf("%d",&n))        printf("%d\n",dp[n]);    return 0;}

法二：母函数：

# include <stdio.h># define MAXN 120int a[MAXN+3]={1}, b[MAXN+3]={0};void init(){    int i, j, k;    for(i=1; i<=MAXN; ++i)//枚举分解的数    {        for(j=0; j<=MAXN; j+=i)//从0开始，步长为i            for(k=0; j+k<=MAXN; ++k)                b[j+k] += a[k];//多项式的系数相加        for(j=0; j<=MAXN; ++j)        {            a[j] = b[j];//更新a数组            b[j] = 0;//清除b数组        }    }}int main(){    init();    int n;    while(~scanf("%d",&n))        printf("%d\n",a[n]);    return 0;}