洛谷P1072Hankson的趣味题题解

题目

一道十分经典的数论题,在考场上也可以用暴力的算法来解决,从而得到\(50pts\)的较为可观的分数,而如果想要AC的话,我们观察原题给的数据范围\(a,b,c,d\)(为了好表示,分别代表a1,a2,b1,b2)。

这样我们可以根据比较容易推出的定理来优化

\[gcd(a,b)==c~=>~gcd(a/c,b/c)==1 \]

\(Code\)

#include <iostream>
#include <bits/stdc++.h>
#define ll long long
using namespace std;
ll __gcd(ll n, ll m)
{
	if (m == 0) return n;
	return __gcd(m, n % m);
}
ll __lcm(ll n, ll m) {return (n * m / __gcd(n, m));}
ll T; ll a, b, c, d;
inline ll solve(ll a, ll b, ll c, ll d)
{
    ll ans = 0; 
    ll n = (a / b), m = (d / c);
    for (int i = 1; i <= (int) sqrt(d); i++)
    {
    	if (d % i != 0) continue;
        if (i % b == 0 && __gcd(i / b, n) == 1 && __gcd(m, d / i) == 1) ans++;
        if (d / i == i) continue; 
        ll ha = d / i;
        if (ha % b == 0 && __gcd(ha / b, n) == 1 && __gcd(m, d / ha) == 1) ans++;
    }    
    return ans;
}
int main()
{
    scanf("%lld", &T);
    while (T--)
    {
        scanf("%lld%lld%lld%lld", &a, &b, &c, &d);
        printf("%lld\n", solve(a, b, c, d));
    }
}
posted @ 2019-03-14 17:15  DAGGGGGGGGGGGG  阅读(139)  评论(0编辑  收藏  举报