Poj3358: Euler's Theorem

It is also a question which is determined by Euler .. Well, the key is to build an equation and pay attention to some simplification techniques.

Question:

Given a cyclic decimal number generated by P/Q, find the minimum cyclic section in the binary representation of this cyclic decimal number and not the prefix of the cyclic section.

Ideas:

Decimal to binary. We should take the remainder by 2 and set the cycle section with the length of Y starting from the decimal point X,

First, convert P/Q to the simplest score. At this time, p and q are mutually qualitative.

Then it must satisfy the same remainder equation p * 2 ^ x = p * 2 ^ (x + y) mod q

Sort out q | p * 2 ^ x * (2 ^ y-1) q | 2 ^ x * (2 ^ y-1) due to the mutual quality of p and q)

At this time, because 2 ^ Y-1 is an odd number, then there is a division formula we can know that the number of the factor of 2 in Q is X, so we can deal with Q to get X, change Q to Q/(2 ^ X );

Finally, we can obtain the homogeneous equation 2 ^ y = 1 (mod q)

Use Euler's theory to understand this same-remainder equation.

The Code is as follows:

#include <iostream>#include <stdio.h>#include<string.h>#include<algorithm>#include<string>#include<ctype.h>using namespace std;#define MAXN 10000long long gcd(long long a,long long b){    return b?gcd(b,a%b):a;}long long phi(long long n){    long long res=n;    for(int i=2;i*i<=n;i++)    {        if(n%i==0)        {            res=res-res/i;            while(n%i==0)            {                n/=i;            }        }    }    if(n>1)        res=res-res/n;    return res;}long long multi(long long a,long long b,long long m)//a*b%m{    long long res=0;    while(b>0)    {        if(b&1)            res=(res+a)%m;        b>>=1;        a=(a<<1)%m;    }    return res;}long long quickmod(long long a,long long b,long long m) //a^b%m{    long long res=1;    while(b>0)    {        if(b&1)            res=multi(res,a,m);        b>>=1;        a=multi(a,a,m);    }    return res;}int main(){    long long p,q,x,y;    int cas=0;    while(scanf("%I64d/%I64d",&p,&q)!=EOF)    {        if(p==0)        {            puts("1,1");            continue;        }        cas++;        long long t=gcd(p,q);        x=1;        p/=t;q/=t;        while(q%2==0)        {            q/=2;x++;        }        long long m=phi(q);        y=m;        for(long long i=2;i*i<=m;i++)        {            if(m%i==0)            {                while(m%i==0)                    m/=i;                while(y%i==0)                {                    y/=i;                    if(quickmod(2,y,q)!=1)                    {                        y*=i;                        break;                    }                }            }        }        printf("Case #%d: %I64d,%I64d\n",cas,x,y);    }    return 0;}

 

Poj3358: Euler's Theorem