Project Euler 141

题目

Investigating progressive numbers, n, which are also square

A positive integer, \(n\), is divided by \(d\) and the quotient and remainder are \(q\) and \(r\) respectively. In addition \(d, q\), and \(r\) are consecutive positive integer terms in a geometric sequence, but not necessarily in that order.

For example, \(58\) divided by \(6\) has quotient \(9\) and remainder \(4\). It can also be seen that \(4, 6, 9\) are consecutive terms in a geometric sequence (common ratio \(\dfrac{3}{2}\)).

We will call such numbers, \(n\), progressive.

Some progressive numbers, such as \(9\) and \(10404 = 102^2\), happen to also be perfect squares.

The sum of all progressive perfect squares below one hundred thousand is \(124657\).

Find the sum of all progressive perfect squares below one trillion (\(10^{12}\)).

解决方案

按照题意，有等式\(n=qd+r\)，其中\(r< q\)是可以确定的.

那么假设\(d< q\)（因为无论\(q<d\)还是\(q>d\)，对上面\(n\)的结果不会有任何改变。），并且这个公比是一个有理数，由最简分数\(\dfrac{a}{b}\)表示，那么有

\(d=\dfrac{a}{b} \cdot r,q=\dfrac{a^2}{b^2}\cdot r\)

如果\(q\)是一个正整数，那么\(b^2\mid r\)。因此就存在一个\(c\)，使得\(r=cb^2\)。

此时\(d=abc,q=a^2c\)。

那么\(n=qd+r=a^3bc^2+b^2c\)。

枚举这三个数\(a,b,c\)即可，要保证\(\gcd(a,b)=1,a>b\)。

代码

# include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll N=1e12;
unordered_set<ll>st1,st;
int main(){
    for(ll i=1;i*i<N;i++)
        st1.insert(i*i);
    for(ll a=2,a3;(a3=a*a*a)<N;a++){
        ll d=2-a%2;
        for(ll b=1;b<a&&b*(a3+b)<N;b+=d){
            if(__gcd(a,b)==1){
                for(ll c=1,w;(w=b*c*(b+a3*c))<N;c++){
                    if(st1.count(w))
                        st.insert(w);
                }
            }
        }
    }
    ll ans=0;
    printf("%lld\n",ans);
}