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POJ 2429 GCD & LCM Inverse (大数分解)

时间:2017-05-29 11:54:43      阅读:206      评论:0      收藏:0      [点我收藏+]

标签:tmp   target   art   amp   str   while   sort   sizeof   problem   

GCD & LCM Inverse

题意:
给你两个数的gcd和lcm,[1, 2^63)。

求a,b。使得a+b最小。


思路:
lcm = a * b / gcd 将lcm/gcd之后进行大数分解。形成a^x1 * b^x2 * c^x3…… 的形式。当中a,b,c为互不同样的质数。然后暴力枚举就可以。

代码:
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<cmath>
#include<cstdio>
#include<vector>
#include<string>
#include<fstream>
#include<cstring>
#include<ctype.h>
#include<iostream>
#include<algorithm>
#define INF (1<<30)
#define PI acos(-1.0)
#define mem(a, b) memset(a, b, sizeof(a))
#define rep(i, n) for (int i = 0; i < n; i++)
#define debug puts("===============")
typedef long long ll;
using namespace std;
const int S = 20; 

//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 <2^63
ll mult_mod(ll a, ll b, ll c) {
    a %= c;
    b %= c;
    ll 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 %c
ll pow_mod(ll x, ll n, ll mod) { //x^n%c
    if(n == 1)return x % mod;
    x %= mod;
    ll tmp = x;
    ll ret = 1;
    while(n) {
        if(n & 1) ret = mult_mod(ret, tmp, mod);
        tmp = mult_mod(tmp, tmp, mod);
        n >>= 1;
    }
    return ret;
}

//以a为基,n-1=x*2^t      a^(n-1)=1(mod n)  验证n是不是合数
//一定是合数返回true,不一定返回false
bool check(ll a, ll n, ll x, ll t) {
    ll ret = pow_mod(a, x, n);
    ll last = ret;
    for(int i = 1; i <= t; i++) {
        ret = mult_mod(ret, ret, n);
        if(ret == 1 && last != 1 && last != n - 1) return true; //合数
        last = ret;
    }
    if(ret != 1) return true;
    return false;
}

// Miller_Rabin()算法素数判定
//是素数返回true.(可能是伪素数,但概率极小)
//合数返回false;
bool Miller_Rabbin(ll n) {
    if(n < 2)return false;
    if(n == 2)return true;
    if((n & 1) == 0) return false; 
    ll x = n - 1;
    ll t = 0;
    while((x & 1) == 0) {
        x >>= 1;
        t++;
    }
    for(int i = 0; i < S; i++) {
        ll a = rand() % (n - 1) + 1; 
        if(check(a, n, x, t))
            return false;//合数
    }
    return true;
}

//************************************************
//pollard_rho 算法进行质因数分解
//************************************************
ll factor[100];//质因数分解结果(刚返回时是无序的)
int tot;//质因数的个数。

数组小标从0開始 ll gcd(ll a, ll b) { if(a == 0) return 1; if(a < 0) return gcd(-a, b); while(b) { ll t = a % b; a = b; b = t; } return a; } ll Pollard_rho(ll x, ll c) { ll i = 1, k = 2; ll x0 = rand() % x; ll y = x0; while(1) { i++; x0 = (mult_mod(x0, x0, x) + c) % x; ll d = gcd(y - x0, x); if(d != 1 && d != x) return d; if(y == x0) return x; if(i == k) { y = x0; k += k; } } } //对n进行素因子分解 void findfac(ll n) { if(Miller_Rabbin(n)) { //素数 factor[tot++] = n; return; } ll p = n; while(p >= n) p = Pollard_rho(p, rand() % (n - 1) + 1); findfac(p); findfac(n / p); } int vis[111] = {0}; ll g, lcm; ll ans = (1LL << 61) , s; int main () { while(~scanf("%lld%lld", &g, &lcm)) { lcm /= g; if (lcm == 1) { printf("%lld %lld\n", g, g); continue; } tot = 0; findfac(lcm); sort(factor, factor + tot); int cnt = 0; ll a[111]; ll res = 1, now = factor[0]; for (int i = 0; i < tot; i++) { if (factor[i] == now) { res = res * now; } else { a[cnt++] = res; now = factor[i]; res = now; } } a[cnt++] = res; int t = (1 << cnt); ll A = 1, B = lcm; for (int i = 0; i < t; i++) { ll TA = 1, TB = 1; for (int j = 0; j < cnt; j++) if (i & (1 << j)) TA *= a[j]; else TB *= a[j]; if (TA + TB < A + B) { A = TA; B = TB; } } if (A > B) swap(A, B); printf("%lld %lld\n", A * g, B * g); } return 0; }



POJ 2429 GCD &amp; LCM Inverse (大数分解)

标签:tmp   target   art   amp   str   while   sort   sizeof   problem   

原文地址:http://www.cnblogs.com/ljbguanli/p/6917611.html

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