joj 1026 动态规划，生成函数及改进

joj 1026 动态规划，生成函数及改进

One curious child has a set of N little bricks. From these bricks he builds different staircases. Staircase consists of steps of different sizes in a strictly descending order. It is not allowed for staircase to have steps equal sizes. Every staircase consists of at least two steps and each step contains at least one brick. Picture gives examples of staircase for N=11 and N=5:

Your task is to write a program that reads from input numbers N and writes to output numbers Q - amount of different staircases that can be built from exactly N bricks.

Input

Numbers N, one on each line. You can assume N is between 3 and 500, both inclusive. A number 0 indicates the end of input.

 

 

Output

Numbers Q, one on each line.

 

Sample Input

3
5
0

Sample Output

1
2

方法1，动态规划

#include<iostream>
#include<cstdlib>
using namespace std;
  int main()
  {
  freopen("s.txt","r",stdin);
  freopen("key.txt","w",stdout);
  double f[501][501]={0};
  double s;
  int i,j,k,n;
  for(i=3;i<=500;i++)
  for(j=1;j<=(i-1)/2;j++)
    f[i][j]=1;
  for(i=3;i<=500;i++)
    for(j=1;j<=(i-1)/2;j++)
    {
 for(k=j+1;k<=(i-j-1)/2;k++)
   f[i][j]=f[i-j][k]; 
 }
    while(scanf("%d",&n),n) {
        s=0;
        for(i=1;i<=(n-1)/2;i++) s+=f[n][i]; //f[n]=f[n][1]+f[n][2]+-----+f[n][floor((i-1)/2)]
        printf("%.0f\n",s);
    }
  //system("PAUSE");
  return   0;
  }
更妙的方法：生成函数法
计算(1+x)(1+x^2)(1+x^3)-----,x^n的系数即为所求
int i,j;
double ans[510]={1,1};//已经把ans[1]和ans[0]赋为1了，其余为0
 for(i=2;i<=500;i++) {  
        for(j=500;j>=0;j--) { 
            if(i+j<=500) ans[i+j]+=ans[j];
        } 
    } 

先计算(1+x)(1+x^2)
再计算(1+x)(1+x^2)   *(1+x^3)