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FactorialSystem (code jam china round2 300分真題)

Problem Statement

In the factorial number system the value of the first digit (from the right) is 1!, the value of the second digit is 2!, ..., and the value of the n-th digit is n!. This means that any decimal number d can be written in this system as: a_na_n-1...a₂a₁, where

d = a_n * n! + a_n-1 * (n-1)! + ... + a₂ * 2! + a₁ * 1! and 0 <= a_i <= i for all i.

Given an int num in decimal, return its representation in the factorial number system.

Definition

Class:	FactorialSystem
Method:	convert
Parameters:	int
Returns:	int
Method signature:	int convert(int num)
(be sure your method is public)

Notes

n! = 1 * 2 * ... * n

Constraints

num will be between 1 and 3628799 (10!-1), inclusive.

Examples

Returns: 1

Returns: 1000

24 = 4!

Returns: 11111

153 = 1! + 2! + 3! + 4! + 5!.

Returns: 10201

Returns: 987654321

Largest possible input.

This problem statement is the exclusive and proprietary property of TopCoder, Inc. Any unauthorized use or reproduction of this information without the prior written consent of TopCoder, Inc. is strictly prohibited. (c)2003, TopCoder, Inc. All rights reserved.

posted on 2005-12-23 10:26 emu 閱讀(1349) 評論(4) 編輯收藏所屬分類: google編程大賽模擬題及入圍賽真題

package factorial;
import java.util.LinkedList;
public class Factorial {

public static void main(String[] args) {
StringBuffer sb=new StringBuffer();
LinkedList<Integer> al=new LinkedList<Integer>();
int m=445;
int z=1;
int fZ=z;
al.add(fZ);
while(m/fZ>0)
{
z++;
al.addLast(al.getLast()*z);
fZ=al.getLast();

}
int n=z-1;
for(int i=n;i>=1;i--)
{

int temp=m/al.get(i-1);
m-=temp*al.get(i-1);
sb.append(temp);
}
System.out.println(sb.toString());

}

}
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# re: FactorialSystem (code jam china round2 300分真題) 2008-10-12 21:45 zgqchina

#include<iostream>
using namespace std;

long factrial(int n) //求階乘,n>0
{
if(n==1) return 1;
else
return n*factrial(n-1);
}

int k=0,a[9]; //全局變量 K用來計數(shù)，數(shù)組用來存放數(shù)列

void convert(double num) //轉(zhuǎn)換成數(shù)列
{
int n=9;
while((num>0)&&(n>=1))
{
int m=factrial(n);
if(num<m)
a[k]=0;
else
{
for(int p=n;p>=1;p--)
if(num>=p*m)
{
a[k]=p;
break;
}
num=num-p*m;
}
n--;
k++;
}
}

void main()
{
cout<<"please input a number(num will be between 1 and 3628799) :"<<endl;
double temp;
cin>>temp;
cout<<"now the result is :"<<endl;
convert(temp);
int i=0;
while(a[i]==0) i++;
for(i;i<9;i++)
cout<<a[i]<<" ";
} 回復更多評論

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