HDU 1085 Ignatius and the Princess III （母函数-整数拆分）

Ignatius and the Princess III

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11780    Accepted Submission(s): 8340

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
   
   
   
   

 

题意：就是120以内的数进行整数拆分，问总共有多少拆分方法。

如 4可以有5种方式：

4 = 4;

 4 = 3 + 1;
  4 = 2 + 2;
  4 = 2 + 1 + 1;
  4 = 1 + 1 + 1 + 1;   

总共有5种，所以输出5

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>

using namespace std;

const int MAX = 140;

int ans[MAX],tmp[MAX];

void work(int num){
    int i,j,k;

    memset(ans,0,sizeof(ans));
    memset(tmp,0,sizeof(tmp));

    for(i=0;i<=num;++i){
        ans[i] = 1;
    }

    //(author : CSDN iaccepted)
    for(i=2;i<=num;++i){
        for(j=0;j<=num;++j){
            for(k=0;k<=num && j+k*i<=num;++k){
                tmp[j+k*i] += ans[j];
            }
        }
        for(j=0;j<=num;++j){
            ans[j] = tmp[j];
            tmp[j] = 0;
        }
    }
}

int main(){
    //freopen("in.txt","r",stdin);
    //(author : CSDN iaccepted)
    int num;
	work(120);
    while(scanf("%d",&num)!=EOF){
        printf("%d\n",ans[num]);
    }
    return 0;
}