标签:大数分解
#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开始 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 & LCM Inverse (大数分解),布布扣,bubuko.com
POJ 2429 GCD & LCM Inverse (大数分解)
标签:大数分解
原文地址:http://blog.csdn.net/sio__five/article/details/38323605