【hiho一下 第144周】机会渺茫

【题目链接】:http://hihocoder.com/contest/hiho144/problem/1

【题意】

【题解】



找出两个数相同的因子的个数x

然后两个数各自的因子的个数numa,nub;

x/(numa*numb)就是答案了；

统计相同因子个数的时候可以用个map;

因子个数不会像你想得那么多的；

很少的。



【完整代码】

#include <bits/stdc++.h>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define mp make_pair
#define ps push_back
#define fi first
#define se second
#define rei(x) scanf("%d",&x)
#define rel(x) scanf("%lld",&x)
#define ref(x) scanf("%lf",&x)
typedef pair<int, int> pii;
typedef pair<LL, LL> pll;
const int dx[9] = { 0,1,-1,0,0,-1,-1,1,1 };
const int dy[9] = { 0,0,0,-1,1,-1,1,-1,1 };
const double pi = acos(-1.0);
const int N = 110;
LL n, m,numa,numb,ya,yb;
map <LL, int> dic;
LL gcd(LL a, LL b){
    if (b == 0) return a;else return gcd(b, a%b);
}
int main(){
    //freopen("F:\\rush.txt", "r", stdin);
    rel(n), rel(m);
    LL i,len = sqrt(n);
    for (i = 1; i <= len; i++)
        if (n%i == 0){
            if (i*i == n) ya++; else ya += 2;
            dic[i] = dic[n / i] = 1;
        }
    len = sqrt(m);
    for (i = 1;i <= len;i++)
        if (m%i == 0){
            if (i*i == m)
                numa += dic[i],yb++;
            else
                numa += dic[i] + dic[m / i],yb+=2;
        }
    numb = ya*yb;
    LL t = gcd(numa, numb);
    numa /= t, numb /= t;
    printf("%lld %lld\n", numb, numa);
    //printf("\n%.2lf sec \n", (double)clock() / CLOCKS_PER_SEC);
    return 0;
}