stuck

没做出来,先记录一下思路.

找 $a+b+c\le 1e7$ 的所有正整数解

首先令苹果=a,香蕉=b,菠萝=c,化简式子得到

令

再次化简得到

ref,可能要用椭圆曲线

https://gemini.google.com/app/3a778d7e16bdf60e

WA

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#include <iostream>
#include <cmath>
#include <numeric>
#include <set>
#include <vector>
#include <algorithm>

using namespace std;

typedef long long ll;

// 快速最大公约数
ll gcd(ll a, ll b) {
    while (b) { a %= b; swap(a, b); }
    return a;
}

long long res=0;
int main() {
    ll limit = 10000000;
    ll total_cnt = 0;
    // 存储原始解 (a0, b0, c0)
    set<vector<ll>> primitives;

    // 预计算边界
    const double s21 = sqrt(21.0);
    const double s27 = sqrt(27.0);

    for (ll v = 1; v <= 3500; ++v) {
        ll u_min = ceil(s21 * v);
        ll u_max = floor(s27 * v);

        for (ll u = u_min; u <= u_max; ++u) {
            if (gcd(u, v) != 1) continue;

            // 共有项
            ll b_base = 2 * (27 * v * v - u * u);
            ll c_base = 3 * (u * u - 21 * v * v);

            // 方案 1: 对应二次方程较大的根
            auto add_sol = [&](ll a, ll b, ll c) {
                if (a <= 0 || b <= 0 || c <= 0) return;
                ll g = gcd(a, gcd(b, c));
                primitives.insert({a / g, b / g, c / g});
            };

            add_sol(3 * (u * u + 2 * u * v - 15 * v * v), b_base, c_base);

            // 方案 2: 对应二次方程较小的根 (当 u > 5v 时 x2 > 0)
            if (u > 5 * v) {
                add_sol(3 * (u * u - 2 * u * v - 15 * v * v), b_base, c_base);
            }
        }
    }

    // 计算所有倍数解
    for (auto& sol : primitives) {
        ll sum = sol[0] + sol[1] + sol[2];
        if (sum <= limit) {
            total_cnt ++;
            res+=sum;
        }
    }

    cout << "Total Solutions: " << total_cnt<<" "<<res << endl;

    return 0;
}