Ignatius and the Princess III（母函数一种技巧性的暴力）

Ignatius and the Princess III

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 13070    Accepted Submission(s): 9236

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

 
题意：整数划分，给你一个整数n，问有多少种方式划分整数；比如2可以划分为2,1+1；

意解：母函数模板；  母函数其实就是把组合加法与幂指数的乘法法则联合在一起

         

#include <iostream>#include <cstring>#include <cstdio>using namespace std;const int M = 1e4 * 2;typedef long long ll;ll a[M],b[M];/********//* * 母函数，a数组是维护当前能组成的数的方法数，b数组为中间量； * 母函数其实就是把组合加法于幂指数的乘法法则联合在一起 * 这相当于暴力查找符合题意的方法种数，取于不取的问题，取几个的问题； */ int main(){   int n;   while(~scanf("%d",&n))   {       for(int i = 0; i <= n; i++)       {           a[i] = 1;           b[i] = 0;       }       for(int i = 2; i <= n; i++)       {           for(int j = 0; j <= n; j++) //当前序列的指数           {               for(int k = 0; k + j <= n; k += i) //接下来序列的指数，因为指数是以i递增的；               {                   b[k + j] += a[j];               }           }           for(int j = 0; j <= n; j++) //更新序列；           {               a[j] = b[j];               b[j] = 0;           }       }       printf("%I64d\n",a[n]);   }   return 0;}