POJ 2429 GCD & LCM Inverse(Pollard-rho 大整数分解+DFS)
来源:互联网 发布:微课录制软件下载 编辑:程序博客网 时间:2024/06/06 02:20
GCD & LCM Inverse
Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 16795 Accepted: 3107
Description
Given two positive integers a and b, we can easily calculate the greatest common divisor (GCD) and the least common multiple (LCM) of a and b. But what about the inverse? That is: given GCD and LCM, finding a and b.
Input
The input contains multiple test cases, each of which contains two positive integers, the GCD and the LCM. You can assume that these two numbers are both less than 2^63.
Output
For each test case, output a and b in ascending order. If there are multiple solutions, output the pair with smallest a + b.
Sample Input
3 60
Sample Output
12 15
Source
POJ Achilles
[Submit] [Go Back] [Status] [Discuss]
Home Page Go Back To top
题意:给你两个数的gcd和lcm,让你求这两个数
题解:我们考虑lcm/gcd,这样得出的两个数一定是互质的。
然后我们考虑分解质因数,两个互质的数一定没有公因数,因此dfs暴力其中一个数的因数即可。
数很大,无法直接暴力分解因数,这里采用Pollard-rho算法分解(不知道的自行百度)
然后怎么保证a+b最小呢,其实两个数最接近时,这两个数的和一定最小(可以证明的)。
呢我们直接dfs即可。
#include<set> #include<map> #include<stack> #include<queue> #include<vector> #include<string> #include<time.h> #include<math.h> #include<stdio.h> #include<iostream> #include<string.h> #include<stdlib.h> #include<algorithm> #include<functional> using namespace std; #define ll long long #define inf 1000000000 #define Mod 10007 #define maxn 50500 #define lowbit(x) (x&-x) #define eps 1e-9 ll cnt, fat[101],mx,ans;ll pri[2550005], a[2550005] = {1,1};ll Multi(ll a, ll b, ll mod) //和上面一样,但比上面慢很多 { ll ans = 0; a %= mod; while(b) { if(b%2==1) ans = (ans+a)%mod, b--; else a = (a+a)%mod, b /= 2; } return ans; } ll Pow(ll a, ll b, ll mod) { ll ans = 1; a %= mod; while(b) { if(b&1) ans = Multi(ans, a, mod), b--; else a = Multi(a, a, mod), b /= 2; } return ans; } ll Gcd(ll a, ll b) { if(b==0)return a;return Gcd(b, a%b);} int Miller_Rabin(ll n) { int i, j, k; ll a, x, y, mod; if(n==2) return 1; if(n<2 || n%2==0) return 0; k = 0, mod = n-1; while(mod%2==0) { k++; mod /= 2; } for(i=1;i<=15;i++) { a = rand()%(n-1)+1; x = Pow(a, mod, n); y = 0; for(j=1;j<=k;j++) { y = Multi(x, x, n); if(y==1 && x!=1 && x!=n-1) return 0; x = y; } if(y!=1) return 0; } return 1; } ll Divi(ll n) { ll i, k, x, y, p, c; if(n==1) return 1; k = 2, p = 1; y = x = rand()%n, c = rand()%(n-1)+1; for(i=1;p==1;i++) { x = (Multi(x, x, n)+c)%n;p = x-y;if(p<0)p = -p; p = Gcd(n, p); if(i==k) y = x, k *= 2; } return p; } void Pollard_rho(ll n) { ll p; if(n==1) return; if(Miller_Rabin(n)) fat[++cnt] = n; else { p = Divi(n);Pollard_rho(p); Pollard_rho(n/p); } } void dfs(ll x,ll y){if(x>cnt){if(y>ans && y<=mx)ans=y;return;}dfs(x+1,y);dfs(x+1,y*fat[x]);}int main(void){ll a,b;while(scanf("%lld%lld",&a,&b)!=EOF){cnt=0;b=b/a;Pollard_rho(b);sort(fat+1,fat+cnt+1);int i,j=1;for(i=2;i<=cnt;i++){while(fat[i]==fat[i-1] && i<=cnt)fat[j]*=fat[i],i++; if(i<=cnt)fat[++j]=fat[i];} cnt=j;ans=1;mx=(ll)sqrt((double)b);dfs(1ll,1ll);printf("%lld %lld\n",ans*a,b/ans*a);}return 0;}
阅读全文
0 0
- POJ 2429 GCD & LCM Inverse(Pollard-rho 大整数分解+DFS)
- POJ 2429 GCD & LCM Inverse(素数判定Miller-Rabin+素因子分解Pollard-rho)
- 【POJ2429】GCD & LCM Inverse-Pollard-rho分解+枚举
- poj 2429 GCD & LCM Inverse(大数质因数分解+DFS)
- POJ 2429 GCD & LCM Inverse【miller-rabin算法+pollard rho 算法】
- POJ 2429 GCD & LCM Inverse(大整数素因子分解+二进制枚举)
- poj 2429 GCD & LCM Inverse (pollard
- Pollard Rho 大整数分解
- Pollard-rho大整数分解
- POJ2429 GCD & LCM Inverse (大整数分解)
- POJ 2429 GCD & LCM Inverse (整数分解,由gcd+lcm求a,b)
- poj 2429 GCD & LCM Inverse(拉宾米勒测试+大数分解+dfs)
- POJ 2429 GCD & LCM Inverse (大数分解)
- POJ 2429GCD & LCM Inverse(大数分解)
- POJ 2429 GCD & LCM Inverse(Pollard_rho法质因数分解)
- POJ2429 GCD & LCM Inverse pollard_rho大整数分解
- POJ 2429 GCD & LCM Inverse (大整数素性测试与因式分解)
- Miller-Rabin素数测试和Pollard-rho大整数分解
- 《MFQ&PPDCS》学习心得--TCO(测试覆盖大纲)
- 抽象类、最终类和接口
- 精通棋牌游戏开发
- web.xml 中spring-servlet.xml 和 application.xml 配置位置及含义
- cURL error 56: Rev failure: Connection was reset (see http://curl.haxx.se/libcurl/c/libcurl-errors.h
- POJ 2429 GCD & LCM Inverse(Pollard-rho 大整数分解+DFS)
- Construct2小游戏开发经验
- thinkphp5的入门学习(2)不同数据添加的规则
- 深度学习进阶(二)--神经网络结构算法以及梯度下降法
- RecyclerView的理解
- JSON数组
- Tensorflow-GPU版本安装(Ubuntu14.04LTS+Cuda8+Quadro K1200)
- Objective-C学习笔记(二)
- 设计模式深入浅出(三)接口适配——桥接