HDU-1028 Ignatius and the Princess III

Ignatius and the Princess III

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

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

————————————————————集训20.3的分割线————————————————————

思路：一开始遇见这道题，我以为是整数拆分，当时想用传说中的分治去做，失败了。后来发现是母函数入门。

代码如下：

/*ID: j.sure.1PROG:LANG: C++*//****************************************/#include <cstdio>#include <cstdlib>#include <cstring>#include <algorithm>#include <cmath>#include <stack>#include <queue>#include <vector>#include <map>#include <string>#include <iostream>#define INF 0x3f3f3f3f#define LL long longusing namespace std;/****************************************/const int N = 125;int c1[N], c2[N];int main(){#ifdef J_Sure//    freopen("000.in", "r", stdin);//    freopen(".out", "w", stdout);#endif    int n;    while(~scanf("%d", &n)) {        memset(c1, 0, sizeof(c1));        c1[0] = 1;        for(int i = 1; i <= n; i++) {            memset(c2, 0, sizeof(c2));            for(int j = 0; j * i <= n; j++) {                for(int k = 0; k + j*i <= n; k++) {                    c2[k+j*i] += c1[k];                }            }            memcpy(c1, c2, sizeof(c2));        }        printf("%d\n", c1[n]);    }    return 0;}