球形空间产生器(Spherical Space Generator) 题解

思路分析

按题目要求使用对应数据结构与算法模拟/递推求解，注意数据范围边界与时间复杂度。

参考代码

#include <cmath>
#include <iomanip>
#include <iostream>
#include <vector>
using namespace std;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int n;
    if (!(cin >> n)) return 0;
    vector<vector<double>> p(n + 1, vector<double>(n));
    for (int i = 0; i <= n; ++i) {
        for (int j = 0; j < n; ++j) cin >> p[i][j];
    }
    vector<vector<double>> a(n, vector<double>(n + 1, 0.0));
    for (int i = 0; i < n; ++i) {
        double rhs = 0.0;
        for (int j = 0; j < n; ++j) {
            a[i][j] = 2.0 * (p[i + 1][j] - p[i][j]);
            rhs += p[i + 1][j] * p[i + 1][j] - p[i][j] * p[i][j];
        }
        a[i][n] = rhs;
    }
    for (int col = 0; col < n; ++col) {
        int piv = col;
        for (int i = col + 1; i < n; ++i) {
            if (fabs(a[i][col]) > fabs(a[piv][col])) piv = i;
        }
        swap(a[col], a[piv]);
        for (int i = 0; i < n; ++i) {
            if (i == col) continue;
            double f = a[i][col] / a[col][col];
            for (int j = col; j <= n; ++j) a[i][j] -= f * a[col][j];
        }
    }
    cout << fixed << setprecision(3);
    for (int i = 0; i < n; ++i) {
        double v = a[i][n] / a[i][i];
        if (fabs(v) < 0.0005) v = 0.0;
        cout << v << (i + 1 == n ? "\n" : " ");
    }
    return 0;
}