HDU1028 Ignatius and the Princess III，母函数

母函数的入门题，母函数适合用来计算整数拆分的方案数，详情可以参考点击打开链接

/*******************************************************************************
 # Author : Neo Fung
 # Email : [email protected]
 # Last modified: 2012-06-26 17:08
 # Filename: HDU1028 Ignatius and the Princess III.cpp
 # Description : 
 ******************************************************************************/
#ifdef _MSC_VER
#define DEBUG
#define _CRT_SECURE_NO_DEPRECATE
#endif

#include <fstream>
#include <stdio.h>
#include <iostream>
#include <string.h>
#include <string>
#include <limits.h>
#include <algorithm>
#include <math.h>
#include <numeric>
#include <functional>
#include <ctype.h>
using namespace std;

const int kMAX=121;
const double kEPS=10E-6;

int tans[kMAX],ans[kMAX];

int main(void)
{
#ifdef DEBUG  
  freopen("../stdin.txt","r",stdin);
  freopen("../stdout.txt","w",stdout); 
#endif  

  int n;

	memset(tans,0,sizeof(tans));
	fill(ans,ans+kMAX,1);

	for(int i=2;i<kMAX;++i)
	{
		for(int j=0;j<kMAX;++j)
			for(int k=0;k+j<kMAX;k+=i)
				tans[k+j]+=ans[j];
		memcpy(ans,tans,sizeof(tans));
		memset(tans,0,sizeof(tans));
	}

  while(~scanf("%d",&n) && n)
  {
		printf("%d\n",ans[n]);
  }

  return 0;
}