poj_3358_Period of an Infinite Binary Expansion

Period of an Infinite Binary Expansion

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 2564 Accepted: 673

Description

Let {x} = 0.a₁a₂a₃... be the binary representation of the fractional part of a rational number z. Suppose that {x} is periodic then, we can write

{x} = 0.a₁a₂...a_r(a_r+1a_r+2...a_r+s)^w

for some integers r and s with r ≥ 0 and s > 0. Also, (a_r+1a_r+2...a_r+s)^wdenotes a nonterminating and repeating binary subsequence of {x}.

The subsequence x₁ = a₁a₂ ... a_ris called the preperiod of {x} and x₂ = a_r+1a_r+2 ... a_r+s is the period of {x}.

Suppose that |x₁| and |x₂| are chosen as small as possible then x₁ is called the least preperiod and x₂ is called the least period of {x}.

For example, x = 1/10 = 0.0001100110011(00110011)^w and 0001100110011 is a preperiod and 00110011 is a period of 1/10.

However, we can write 1/10 also as 1/10 = 0.0(0011)^w and 0 is the least preperiod and 0011 is the least period of 1/10.

The least period of 1/10 starts at the 2nd bit to the right of the binary point and the the length of the least period is 4.

Write a program that finds the position of the first bit of the least period and the length of the least period where the preperiod is also the minimum of a positive rational number less than 1.

Input

Each line is test case. It represents a rational number p/q where p and q are integers, p ≥ 0 and q > 0.

Output

Each line corresponds to a single test case. It represents a pair where the first number is the position of the first bit of the least period and the the second number is the length of the least period of the rational number.

Sample Input

1/10 1/5 101/120 121/1472

Sample Output

Case #1: 2,4 Case #2: 1,4 Case #3: 4,4 Case #4: 7,11

Source

Manila 2006

题型：数论（欧拉函数+费马小定理）

题意：求分数的二进制的循环节的起始位置和循环长度。

分析：

       可以观察一下1/10这组数据，按照二进制转换法(乘二法就是*2取整)，可以得到：
                                                     1/10 2/10 4/10 8/10 16/10 32/10 ...
       然后都分子都尽可能减去10，得到：
                                                     1/10 2/10 4/10 8/10 6/10 2/10 ...
       这时候，发现出现了重复，那么这个重复就是要求的最小循环。

       首先处理一下分数，即将p/q化简，p=gcd(p,q) , q=gcd(p,q)。

       然后将q不断除2知道不能除尽为止，除的次数+1就是循环节的起始位置。

       下一步就是求2^x = 1 (mod q)的最小的x，如何求呢？

        枚举phi（q）的因子，判断2^x = 1 (mod q)，取最小的x。

代码：

#include<iostream>#include<cstring>#include<cstdio>#include<cmath>#define LL long longusing namespace std;LL gcd(LL a, LL b) {    return b ? gcd(b, a % b) : a;}//计算 (a*b)%c.   a,b都是long long的数，直接相乘可能溢出的//  a,b,c <2^63long long mult_mod(long long a,long long b,long long c){    a%=c;    b%=c;    long long ret=0;    while(b)    {        if(b&1){ret+=a;ret%=c;}        a<<=1;        if(a>=c)a%=c;        b>>=1;    }    return ret;}//计算  x^n %clong long pow_mod(long long x,long long n,long long mod)//x^n%c{    if(n==1)return x%mod;    x%=mod;    long long tmp=x;    long long ret=1;    while(n)    {        if(n&1) ret=mult_mod(ret,tmp,mod);        tmp=mult_mod(tmp,tmp,mod);        n>>=1;    }    return ret;}LL phi(LL n){    LL ret = n;    for(LL i=2;i*i<=n;i+=(i==2)?1:2){        if(!(n%i)){            ret=ret/i*(i-1);            while(n%i==0) n/=i;        }    }    if(n>1) ret=ret/n*(n-1);    return ret;}int main(){    //getphi();    LL p,q,time=1;    while(~scanf("%I64d/%I64d",&p,&q)){        LL temp=gcd(p,q);        p/=temp;        q/=temp;        LL i=1;        while(q%2==0){            q/=2;            i++;        }        LL a=phi(q);        LL j=-1;        LL k;        for(k=2;k*k<=a;k++){            if(a%k==0){                if(pow_mod(2,k,q)==1){                    if(j==-1){                        j=k;                    }                    else if(j>k){                        j=k;                    }                    break;                }                else{                    LL temp=a/k;                    if(pow_mod(2,temp,q)==1){                        if(j==-1){                            j=temp;                        }                        else if(j>temp){                            j=temp;                        }                    }                }            }        }        if(j==-1)            j=a;        printf("Case #%I64d: %I64d,%I64d\n",time,i,j);        time++;    }    return 0;}