数据范围很大,用米勒罗宾测试和Pollard_Rho法可以分解大数. 模板在代码中 O.O #include <iostream> #include <cstdio> #include <cstring> #include <cstdlib> #include <cmath> using namespace std; __int64 pri[]= {,,,,,,,,,,};//用小素数表做随机种子避免第一类卡米歇尔数的误判 __int64 mul…

集训队有人提到这个算法,就学习一下,如果用到可以直接贴模板,例题:POJ 1811 转自:http://www.cnblogs.com/kuangbin/archive/2012/08/19/2646396.html 传说中的随机算法. 效率极高. 可以对一个2^63的素数进行判断. 可以分解比较大的数的因子. #include<stdio.h> #include<string.h> #include<stdlib.h> #include<time.h> #…

#include <cstdio> #include <cstring> #include <cmath> #include <ctime> #include <cstdlib> #include <iostream> using namespace std; #define ll long long ; ll ans; ll gcd(ll a,ll b){ ) return gcd(-a , b); ) return a; retu…

大数因数分解Pollard_rho 算法 复杂度o^(1/4) #include <iostream> #include <cstdio> #include <algorithm> #include <cmath> #include <cstring> #include <map> using namespace std; ; ; map<long long, int>m; long long Random( long l…

#include<stdio.h> #include<string.h> #include<stdlib.h> #include<time.h> #include<iostream> #include<algorithm> using namespace std; //**************************************************************** // Miller_Rabin 算法进…

Problem about GCD Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 470    Accepted Submission(s): 77 Problem Description Given integer m. Find multiplication of all 1<=a<=m such gcd(a, m)=1 (cop…

Sample Input 2 5 10 Sample Output Prime 2 模板学习: 判断是否是素数,数据很大,所以用miller,不是的话再用pollard rho分解 miller : 通过费马小定理,若N为素数,a^(N-1) = 1 (mod N), 再利用二次判定: 若x为素数,0<x<p, x*x = 1(mod q) #include <cstdio> #include <cstring> #include <iostream> #i…

题目:PolandBall and Hypothesis A. PolandBall and Hypothesis time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output PolandBall is a young, clever Ball. He is interested in prime numbers. He has stat…

素数判定Miller_Rabin算法详解: http://blog.csdn.net/maxichu/article/details/45458569 大数因数分解Pollard_rho算法详解: http://blog.csdn.net/maxichu/article/details/45459533 然后是参考了kuangbin的模板: http://www.cnblogs.com/kuangbin/archive/2012/08/19/2646396.html 模板如下: //快速乘 (a…

GCDLCM 题目链接(点击) 题目描述 In FZU ACM team, BroterJ and Silchen are good friends, and they often play some interesting games. One day they play a game about GCD and LCM. firstly BrotherJ writes an integer A and Silchen writes an integer B on the paper. The…