POJ 2447 RSA 大数分解+逆元+快速幂

链接：http://poj.org/problem?id=2447

题意：

思路：Pollard_Rho质数分解，得到两个素数因子，P,Q，求出T，E，快速幂即可得M。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <map>
#include <cstdlib>
#include <queue>
#include <stack>
#include <vector>
#include <ctype.h>
#include <algorithm>
#include <string>
#include <set>
#include <ctime>
#define PI acos(-1.0)
#define maxn 10005
#define INF 0x7fffffff
#define eps 1e-8
#define seed 31
typedef long long LL;
typedef unsigned long long ULL;
using namespace std;
LL extend_gcd(LL a, LL b, LL &x, LL &y)
{
    if(b==0)
    {
        x=1;
        y=0;
        return a;
    }
    LL r=extend_gcd(b,a%b,x,y);
    LL t=x;
    x=y;
    y=t-a/b*y;
    return r;
}
LL mul_mod(LL a,LL b,LL n)
{
    a=a%n;
    b=b%n;
    LL s=0;
    while(b)
    {
        if(b&1)
            s=(s+a)%n;
        a=(a<<1)%n;
        b=b>>1;
    }
    return s;
}
LL pow_mod(LL a,LL b,LL n)
{
    a=a%n;
    LL s=1;
    while(b)
    {
        if(b&1)
            s=mul_mod(s,a,n);
        a=mul_mod(a,a,n);
        b=b>>1;
    }
    return s;
}
LL gcd(LL a,LL b)
{
    if(a==0) return 1;
    if(a<0) return gcd(-a,b);
    return b==0?a:gcd(b,a%b);
}
LL inv(LL a,LL m)
{
    LL d,x,y;
    d=extend_gcd(a,m,x,y);
    if (d==1)
    {
        x=(x%m+m)%m;
        return x;
    }
    else return -1;
}
LL Pollard_Rho(LL n)
{
    if(!(n&1)) return 2;
    while(true)
    {
        LL x=(LL)rand()%n;
        if(x<0)
            x=-x;
        LL y=x;
        LL c=(LL)rand()%n;
        if(x<0)
            c=-c;
        if(c==0||c==2) c=1;
        for(int i=1,k=2;; i++)
        {
            x = mul_mod(x,x,n);
            if (x >= c) x -= c;
            else x += n - c ;
            if (x == n) x = 0 ;
            if (x == 0) x = n-1;
            else x --;
            LL d = gcd (x>y ? x-y: y-x, n);
            if (d == n) break ;
            if (d != 1) return d ;
            if (i == k)
            {
                y = x;
                k <<= 1 ;
            }
        }
    }
}
int main()
{
    LL C,E,N;
    while(~scanf("%lld%lld%lld",&C,&E,&N))
    {
        LL aa=Pollard_Rho(N);
        LL T=(aa-1)*(N/aa-1);
        LL D=inv(E,T);
        LL M=pow_mod(C,D,N);
        printf("%lld\n",M);
    }
    return 0;
}

POJ 2447 RSA 大数分解+逆元+快速幂