poj3558 euler

来源：互联网发布：安卓录屏软件哪个好编辑：程序博客网时间：2024/06/07 01:54

转载自：http://blog.csdn.net/code_or_code/article/details/38493609

题目：求 q/p 二进制小数的循环节，起点和长度。

若满足 2^phi[ n ] = 1 (mod n ) 则数 t = phi [ n ] 一定有一个使 2^k=1 (mod n )成立的因子 k

#include<iostream>#include<cstdio>#include<cstring>#include<cmath>#include<vector>#include<algorithm>#include<map>using namespace std;const int maxn = 100000 + 10;#define INF 0x3f3f3f3f#define clr(x,y) memset(x,y,sizeof x )typedef long long ll;#define eps 10e-10const ll Mod = 1000000007;typedef pair<ll, ll> P;ll gcd(ll x,ll y){    return y ? gcd(y,x % y) : x;}ll euler(ll n){    ll ans = n;    for(ll i = 2; i * i <= n; i ++)    {        if(n % i == 0)        {            ans = ans / i * (i - 1);            while(n % i == 0)            {                n /= i;            }        }    }    if(n > 1)        ans = ans / n * (n - 1);    return ans;}ll pow_mod(ll x,ll n,ll mod_val){    ll ans = 1;    ll t = x;    while(n)    {        if(n & 1)            ans = ans * t % mod_val;        n >>= 1;        t = t * t % mod_val;    }    return ans;}ll factor[maxn];int work(ll n){    int len = 0;    for(int i = 2; i * i <= n; i ++)    {        if(n % i == 0)        {            factor[len ++] = i;            factor[len ++] = n / i;        }    }    sort(factor,factor + len);    return len;}int main(){    ll n,m;    int T = 1;    while( ~ scanf("%lld/%lld",&n,&m))    {        ll d = gcd(n,m);        n /= d;        m /= d;        ll ans1 = 0,ans2;        ll t = m;        while(t% 2 == 0)        {            t /= 2;            ans1 ++;        }        ll r = t;        ans2 = euler(r);        int len = work(ans2);        for(int i = 0; i < len; i ++)        {            if(pow_mod(2,factor[i],r) == 1)            {                ans2 = factor[i];                break;            }        }        printf("Case #%d: %lld,%lld \n",T ++,ans1 + 1,ans2);    }    return 0;}

转载自：http://blog.csdn.net/code_or_code/article/details/38493609

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