HDOJ 4814 Golden Radio Base

利用题目中给出的公式和hint可以得到两个有用的公式：
phi^(n) = phi^(n-1)+phi^(n-2)
2*(phi^n) = phi^(n+1)+phi^(n-2)
可以计算出phi^100远大于10^9，所以推测最后得到的phi进制的数整数和小数部分应该不会超过100位，事实表明，50位就能过。
所以最终变成了简单的模拟。

Golden Radio Base

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 363    Accepted Submission(s): 165

Problem Description

Golden ratio base (GRB) is a non-integer positional numeral system that uses the golden ratio (the irrational number (1+√5)/2 ≈ 1.61803399 symbolized by the Greek letter φ) as its base. It is sometimes referred to as base-φ, golden mean base, phi-base, or, phi-nary.

Any non-negative real number can be represented as a base-φ numeral using only the digits 0 and 1, and avoiding the digit sequence "11" – this is called a standard form. A base-φ numeral that includes the digit sequence "11" can always be rewritten in standard form, using the algebraic properties of the base φ — most notably that φ + 1 = φ ² . For instance, 11(φ) = 100(φ). Despite using an irrational number base, when using standard form, all on-negative integers have a unique representation as a terminating (finite) base-φ expansion. The set of numbers which possess a finite base-φ representation is the ring Z[1 + √5/2]; it plays the same role in this numeral systems as dyadic rationals play in binary numbers, providing a possibility to multiply.

Other numbers have standard representations in base-φ, with rational numbers having recurring representations. These representations are unique, except that numbers (mentioned above) with a terminating expansion also have a non-terminating expansion, as they do in base-10; for example, 1=0.99999….

Coach MMM, an Computer Science Professor who is also addicted to Mathematics, is extremely interested in GRB and now ask you for help to write a converter which, given an integer N in base-10, outputs its corresponding form in base-φ.

 

Input

There are multiple test cases. Each line of the input consists of one positive integer which is not larger than 10^9. The number of test cases is less than 10000. Input is terminated by end-of-file.

 

Output

For each test case, output the required answer in a single line. Note that trailing 0s after the decimal point should be wiped. Please see the samples for more details.

 

Sample Input

123610

 

Sample Output

110.01100.011010.000110100.0101
Hint
 

 

Source

2013 Asia Regional Changchun

 

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int base=100;int n;int wei[220];int main(){    while(scanf("%d",&n)!=EOF)    {        memset(wei,0,sizeof(wei));        wei[base]=n;        while(true)        {            bool flag=false;            for(int i=0;i<200;i++)            {                if(wei[i]>1)                {                    int t=wei[i];                    wei[i]=t%2;                    wei[i+1]+=t/2;                    wei[i-2]+=t/2;                    flag=true;                }            }            for(int i=0;i<200;i++)            {                if(wei[i]&&wei[i+1])                {                    int t=min(wei[i],wei[i+1]);                    wei[i]-=t; wei[i+1]-=t;                    wei[i+2]+=t;                    flag=true;                }            }            if(flag==false) break;        }        int st=-1,ed=-1;        for(int i=0;i<202;i++)            if(wei[i]) { st=i; break; }        for(int i=202;i>=0;i--)            if(wei[i]) { ed=i; break; }        for(int i=ed;i>=st;i--)        {            printf("%d",wei[i]);            if(i==base&&st!=i) putchar('.');        }        putchar(10);    }    return 0;}