poj3358 Period of an Infinite Binary Expansion

首先要明确分式的二进制表达方式：

//supposing the fraction is a / b, (a < b && (a, b))
//its binary expression is denoted as : 0.bit[1]bit[2]...
seed[0] = a;
for(int i = 1; ; i++){
    bit[i] = (seed[i - 1] << 1) / b;
    seed[i] = (seed[i - 1] << 1) % b;
}

不妨设其中一个循环节为bit[r]..bit[s]，显然有seed[r - 1] = seed[s] 且bit[r] = bit[s + 1]，其中循环节长度l = （s - r + 1)。

由seed[i] = 2ⁱ⁺¹ % b, 则有 2^r  % b = 2^s+1 % b,  即2^r(2^s-r+1-1） ≡ 0（modb），

记b = 2^t*b1,其中 （b1，2），且令r = t，则 2^l≡ 1（modb1)。

则l|φ（b1），由此枚举b1的约数即可得到周期l。

并且有初始位置p = t + 1 。

http://poj.org/problem?id=3358

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
typedef __int64 LL;
const int maxn = 1e2 + 10;
int prime[maxn], k;
int fac[maxn];
int a, b, B;

int power(int a, int p){
    a %= B;
    int ans = 1;
    while(p){
        if(p & 1) ans = (LL)ans * a % B;
        p >>= 1;
        a = (LL)a * a % B;
    }
    return ans;
}

int gcd(int a, int b) { return !b ? a : gcd(b, a % b); }

void solve(){
    a %= b;
    int d = gcd(a, b);
    a /= d, b /= d;
    int t = 0;
    while(b % 2 == 0) b /= 2, ++t;
    printf("%d,", ++t);
    if(b == 1) { printf("%d
",1); return; }
    k = 0;
    B = b;
    int phi = b;
    int mid = (int)sqrt(b);
    for(int i = 3; i <= mid; i += 2){
        if(b % i == 0){
            prime[k++] = i;
            while(b % i == 0) b /= i;
         }
    }
    if(b != 1) prime[k++] = b;
    for(int i = 0; i < k; i++) phi /= prime[i];
    for(int i = 0; i < k; i++) phi *= (prime[i] - 1);
    k = 0;
    for(int i = 1; i * i <= phi; i++){
        if(phi % i == 0) fac[k++] = i, fac[k++] = phi / i;
    }
    sort(fac, fac + k);
    for(int i = 0; i < k; i++) if(power(2, fac[i]) == 1){
        printf("%d
", fac[i]);
        break;
    }
}

int main(){
    //freopen("in.txt", "r", stdin);
    int kase = 0;
    while(~scanf("%d/%d", &a, &b)){
        printf("Case #%d: ", ++kase);
        solve();
    }
    return 0;
}