D_num
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2046 Accepted Submission(s): 573
Problem Description
Oregon Maple was waiting for Bob When Bob go back home. Oregon Maple asks Bob a problem that as a Positive number N, if there are only four Positive number M makes Gcd(N, M) == M then we called N is a D_num. now, Oregon Maple has some Positive numbers, and if a Positive number N is a D_num , he want to know the four numbers M. But Bob have something to do, so can you help Oregon Maple?
Gcd is Greatest common divisor.
Gcd is Greatest common divisor.
Input
Some cases (case < 100);
Each line have a numeral N(1<=N<10^18)
Each line have a numeral N(1<=N<10^18)
Output
For each N, if N is a D_NUM, then output the four M (if M > 1) which makes Gcd(N, M) = M. output must be Small to large, else output “is not a D_num”.
Sample Input
6 10 9
Sample Output
2 3 6 2 5 10 is not a D_num
Source
Recommend
lcy
就是判断一个long long的数的约数是不是有4个。
用pollard_rho,练习了下模板;
//============================================================================ // Name : HDU3864.cpp // Author : // Version : // Copyright : Your copyright notice // Description : Hello World in C++, Ansi-style //============================================================================ #include <iostream> #include <stdio.h> #include <stdlib.h> #include <algorithm> #include <string.h> #include <time.h> using namespace std; const int S=2; long long mult_mod(long long a,long long b,long long c) { a%=c; b%=c; long long ret=0; while(b) { if(b&1){ret+=a;ret%=c;} a<<=1; if(a>=c)a%=c; b>>=1; } return ret; } long long pow_mod(long long x,long long n,long long mod) { if(n==1)return x%mod; x%=mod; long long tmp=x; long long ret=1; while(n) { if(n&1)ret=mult_mod(ret,tmp,mod); tmp=mult_mod(tmp,tmp,mod); n>>=1; } return ret; } long long check(long long a,long long n,long long x,long long t) { long long ret=pow_mod(a,x,n); long long 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; last=ret; } if(ret!=1)return true; return false; } bool Miller_Rabin(long long n) { if(n<2)return false; if(n==2)return true; if((n&1)==0)return false; long long x=n-1; long long t=0; while((x&1)==0){x>>=1;t++;} for(int i=0;i<S;i++) { long long a=rand()%(n-1)+1; if(check(a,n,x,t)) return false; } return true; } long long factor[100]; int tol; long long gcd(long long a,long long b) { if(a==0)return 1; if(a<0)return gcd(-a,b); while(b) { long long t=a%b; a=b; b=t; } return a; } long long Pollard_rho(long long x,long long c) { long long i=1,k=2; long long x0=rand()%x; long long y=x0; while(1) { i++; x0=(mult_mod(x0,x0,x)+c)%x; long long d=gcd(y-x0,x); if(d!=1&&d!=x)return d; if(y==x0)return x; if(i==k) { y=x0; k+=k; } } } void findfac(long long n) { if(Miller_Rabin(n)) { factor[tol++]=n; return; } long long p=n; while(p>=n)p=Pollard_rho(p,rand()%(n-1)+1); findfac(p); findfac(n/p); } int main() { srand(time(NULL)); long long n; while(scanf("%I64d",&n)==1) { if(n==1) { printf("is not a D_num\n"); continue; } tol=0; findfac(n); if(tol!=2 && tol!=3) { printf("is not a D_num\n"); continue; } sort(factor,factor+tol); if(tol==2) { if(factor[0]!=factor[1]) { printf("%I64d %I64d %I64d\n",factor[0],factor[1],factor[0]*factor[1]); continue; } else { printf("is not a D_num\n"); continue; } } if(tol==3) { if(factor[0]==factor[1]&&factor[1]==factor[2]) { printf("%I64d %I64d %I64d\n",factor[0],factor[0]*factor[1],factor[0]*factor[1]*factor[2]); continue; } else { printf("is not a D_num\n"); continue; } } } return 0; }