UVa 11476 - Factorizing Larget Integers

contents

  1. 1. Problem
  2. 2. Sample Input
  3. 3. Sample Output
  4. 4. Solution

Problem

給你一個正整數 $N$ ($1 < N \le10^{18}$),請你把 $N$ 質因數分解。

注意:大整數分解

Sample Input

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2
3
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3
60
36
10007

Sample Output

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2
3
60 = 2^2 * 3 * 5
36 = 2^2 * 3^2
10007 = 10007

Solution

2015/07/11 第二版,加速三倍,加快模乘法運算,減少模數利用減法。

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#include <bits/stdc++.h>
using namespace std;
#define MAXL (50000>>5)+1
#define GET(x) (mark[x>>5]>>(x&31)&1)
#define SET(x) (mark[x>>5] |= 1<<(x&31))
int mark[MAXL];
int P[50000], Pt = 0;
void sieve() {
register int i, j, k;
SET(1);
int n = 46340;
for (i = 2; i <= n; i++) {
if (!GET(i)) {
for (k = n/i, j = i*k; k >= i; k--, j -= i)
SET(j);
P[Pt++] = i;
}
}
}
long long mul(unsigned long long a, unsigned long long b, unsigned long long mod) {
long long ret = 0;
for (a %= mod, b %= mod; b != 0; b >>= 1, a <<= 1, a = a >= mod ? a - mod : a) {
if (b&1) {
ret += a;
if (ret >= mod) ret -= mod;
}
}
return ret;
}
void exgcd(long long x, long long y, long long &g, long long &a, long long &b) {
if (y == 0)
g = x, a = 1, b = 0;
else
exgcd(y, x%y, g, b, a), b -= (x/y) * a;
}
long long llgcd(long long x, long long y) {
if (x < 0) x = -x;
if (y < 0) y = -y;
if (!x || !y) return x + y;
long long t;
while (x%y)
t = x, x = y, y = t%y;
return y;
}
long long inverse(long long x, long long p) {
long long g, b, r;
exgcd(x, p, g, r, b);
if (g < 0) r = -r;
return (r%p + p)%p;
}
long long mpow(long long x, long long y, long long mod) { // mod < 2^32
long long ret = 1;
while (y) {
if (y&1)
ret = (ret * x)%mod;
y >>= 1, x = (x * x)%mod;
}
return ret % mod;
}
long long mpow2(long long x, long long y, long long mod) {
long long ret = 1;
while (y) {
if (y&1)
ret = mul(ret, x, mod);
y >>= 1, x = mul(x, x, mod);
}
return ret % mod;
}
int isPrime(long long p) { // implements by miller-babin
if (p < 2 || !(p&1)) return 0;
if (p == 2) return 1;
long long q = p-1, a, t;
int k = 0, b = 0;
while (!(q&1)) q >>= 1, k++;
for (int it = 0; it < 2; it++) {
a = rand()%(p-4) + 2;
t = mpow2(a, q, p);
b = (t == 1) || (t == p-1);
for (int i = 1; i < k && !b; i++) {
t = mul(t, t, p);
if (t == p-1)
b = 1;
}
if (b == 0)
return 0;
}
return 1;
}
long long pollard_rho(long long n, long long c) {
long long x = 2, y = 2, i = 1, k = 2, d;
while (true) {
x = (mul(x, x, n) + c);
if (x >= n) x -= n;
d = llgcd(x - y, n);
if (d > 1) return d;
if (++i == k) y = x, k <<= 1;
}
return n;
}
void factorize(int n, vector<long long> &f) {
for (int i = 0; i < Pt && P[i]*P[i] <= n; i++) {
if (n%P[i] == 0) {
while (n%P[i] == 0)
f.push_back(P[i]), n /= P[i];
}
}
if (n != 1) f.push_back(n);
}
void llfactorize(long long n, vector<long long> &f) {
if (n == 1)
return ;
if (n < 1e+9) {
factorize(n, f);
return ;
}
if (isPrime(n)) {
f.push_back(n);
return ;
}
long long d = n;
for (int i = 2; d == n; i++)
d = pollard_rho(n, i);
llfactorize(d, f);
llfactorize(n/d, f);
}
int main() {
sieve();
int testcase;
scanf("%d", &testcase);
while (testcase--) {
long long n;
scanf("%lld", &n);
vector<long long> f;
map<long long, int> r;
llfactorize(n, f);
for (auto &x : f)
r[x]++;
printf("%lld =", n);
for (auto it = r.begin(); it != r.end(); it++) {
if (it != r.begin())
printf(" *");
printf(" %lld", it->first);
if (it->second > 1)
printf("^%d", it->second);
}
puts("");
}
return 0;
}