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)
Total Submission(s): 14069    Accepted Submission(s): 9913

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
输入的数是指数，从1到n每个数都有个数为无限次，其实这里设为个数最多为n个即可，套入母函数的模板很容易求得结果。
#include <iostream>#include <fstream>#include <stdio.h>#include <memory.h>using namespace std;int a[150],b[150];void mother(int i,int num,int n){//i代表第i个元素，第i个元素的数量有num个，需要计算的最大指数n    for(int j=0;j<=n;j++)        for(int k=0;k<=num&&k*i+j<=n;k++)        b[k*i+j]+=a[j];    for(int j=0;j<=n;j++)    {        a[j]=b[j];        b[j]=0;    }}int main(void){    int n;    while(scanf("%d",&n)!=EOF)    {        memset(a,0,sizeof(a));        memset(b,0,sizeof(b));       // for(int i=0;i<=n;i++)          //  a[i]=1;          a[0]=1;        for(int i=1;i<=n;i++)            mother(i,n,n);        printf("%d\n",a[n]);    }    return 0;}