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poj2429:因数分解+搜索

题意:给定gcd(a,b)和lcm(a,b) 求使得a+b最小的 a,b

思路:结合算数基本定理中 gcd lcm的质因子表示形式

把lcm(a,b)质因数分解 以后 通过dfs找到 a+b最小的a b即可

#include <iostream>#include<stdio.h>#include<algorithm>#include<math.h>using namespace std;long long fac[100];int nf;long long a,b;long long x,y;long long mk;long long ans1,ans2;long long gcd(long long a,long long b){    return b?gcd(b,a%b):a;}long long random(long long n){    return (long long)(rand()%(n-1)+1);}long long multimod(long long a,long long b,long long m)//a*b%m{    long long res=0;    while(b>0)    {        if(b&1)            res=(res+a)%m;        b>>=1;        a=(a<<1)%m;    }    return res;}long long quickmod(long long a,long long b,long long m) //a^b%m{    long long res=1;    while(b>0)    {        if(b&1)            res=multimod(res,a,m);        b>>=1;        a=multimod(a,a,m);    }    return res;}int check(long long a,long long n,long long x,long long t){    long long res=quickmod(a,x,n);    long long last=res;    for(int i=1;i<=t;i++)    {        res=multimod(res,res,n);        if(res==1&&last!=1&&last!=n-1) return 1;        last=res;    }    if(res!=1) return 1;    return 0;}int primetest(long long n){    if(n<2)return 0;    if(n==2)return 1;    if((n&1)==0) return 0;    long long x=n-1;    long long t=0;    while((x&1)==0){x>>=1;t++;}    for(int i=0;i<20;i++)    {        long long a=random(n);        if(check(a,n,x,t))            return 0;    }    return 1;}long long pollardrho(long long n,long long c){    long long x,y,d,i,k;    i=1;k=2;    x=random(n);    y=x;    while(1)    {        i++;        x=(multimod(x,x,n)+c)%n;        long long tmp=y-x>=0?y-x:x-y;        d=gcd(tmp,n);        if(d>1&&d<n)            return d;        if(y==x)            return n;        if(i==k)        {            y=x;            k+=k;        }    }}void findfac(long long n){    if(n==1)        return;    if(primetest(n))    {        fac[nf++]=n;        return;    }    long long p=n;    while(p>=n)        p=pollardrho(n,random(n-1));    findfac(p);    findfac(n/p);}void dfs(long long x,long long y,int s,long long pre){    while(fac[s]==pre&&(s<nf))        s++;        //因子判重    if(s==nf)    {        if(x+y<mk)        {            mk=x+y;            ans1=x;            ans2=y;        }        return;    }    long long i=1,j=1;    long long a1=a,b1=b;    while(a1%fac[s]==0)    {        a1/=fac[s];    }    while(b1%fac[s]==0)    {        b1/=fac[s];    }    i=a/a1;    j=b/b1;    dfs(x*i,y*j,s+1,fac[s]);    dfs(x*j,y*i,s+1,fac[s]);    return;}int main(){    while(scanf("%I64d %I64d",&a,&b)!=EOF)    {        nf=0;        findfac(b);        mk=9223372036854775807;        sort(fac,fac+nf);        dfs(1,1,0,-1);        if(ans1>ans2)            swap(ans1,ans2);        printf("%I64d %I64d\n",ans1,ans2);    }}

 

poj2429:因数分解+搜索