POJ-2429 GCD & LCM Inverse

题目：

http://poj.org/problem?id=2429

题意：

给出gcd与lcm求原来两数，有多解时找出使得a+b最小的那个。

思路：

模版题，

lcm/gcd==(a/gcd)*(b/gcd).

所以求出lcm/gcd的所有质因子，自由组合成两组就是答案。

再dfs枚举所有情况找出a+b最小的情况。

代码：

#define N 112345
#define T 10

long long  n,m;
long long  flag,ave,ans,res,len,ans1,ans2;

long long mod_mul(long long a,long long b,long long n)
{
long long res = 0;
while(b)
{
if(b&1)    res = (res + a) % n;
a = (a + a) % n;
b >>= 1;
}
return res;
}
long long mod_exp(long long x,long long k,long long mod)
{
long long ans = 1;
while(k)
{
if(k & 1) ans = mod_mul(ans, x, mod);
x = mod_mul(x, x, mod);
k >>= 1;
}
return ans;
}
bool Miller_Rabin(long long n)
{
if(n == 2 || n == 3 || n == 5 || n == 7 || n == 11)    return true;
if(n == 1 || !(n%2) || !(n%3) || !(n%5) || !(n%7) || !(n%11))    return false;

long long x, pre, u;
int i, j, k = 0;
u = n - 1;

while(!(u&1))
{
k++;
u >>= 1;
}

for(i = 0; i < T; ++i)
{
x = rand()%(n-2) + 2;
if((x%n) == 0)
continue;
x = mod_exp(x, u, n);
pre = x;
for(j = 0; j < k; ++j)
{
x = mod_mul(x,x,n);
if(x == 1 && pre != 1 && pre != n-1)
return false;
pre = x;
}
if(x != 1)
return false;
}
return true;
}

long long factor[1000],factor2[1000];
int sum,sum2;

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=(mod_mul(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 x)
{
if(x==1)return ;
if(Miller_Rabin(x))
{
factor[sum++]=x;
return;
}
long long p=x;
while(p>=x)
p=Pollard_rho(p,rand()%(x-1)+1);
findfac(p);
findfac(x/p);
}
void dfs(int num,long long now)
{
if(sum2==num)
{
if(now&&ans>now+m/n/now)
ans=now+m/n/now,res=now;
return ;
}
dfs(num+1,now);
dfs(num+1,now*factor2[num]);
}
int main()
{
long long i,j,k,kk,t,x,y,z;
srand((long long)time(0));
while(scanf("%lld%lld",&n,&m)!=EOF&&n)
{
x=m/n;
sum=0;
findfac(x);

ans=2e63;
sum2=0;
sort(factor,factor+sum);
factor2[0]=factor[0];
for(int i=0;i<sum;i++)
{
if(factor[i]==factor[i+1])
factor2[sum2]*=factor[i+1];
else
factor2[++sum2]=factor[i+1];
}
dfs(0,1);
if(res*n<m/res)
printf("%lld %lld\n",res*n,m/res);
else
printf("%lld %lld\n",m/res,res*n);
}
return 0;
}