高斯消元 n^3

void gauss()
{
    // 转化成上三角矩阵
    for (int r = 1, c = 1; c <= n; c ++, r ++ )//枚举行
    {
        // 找主元 一列里面找绝对值最大
        int t = r;
        for (int i = r + 1; i <= n; i ++ )
            if (fabs(b[i][c]) > fabs(b[t][c]))
                t = i;

        // 交换 换到最上方
        for (int i = c; i <= n + 1; i ++ ) swap(b[t][i], b[r][i]);
        // 归一化 从后往前 把这一行全部化成1
        for (int i = n + 1; i >= c; i -- ) b[r][i] /= b[r][c];
        // 消 用这一行把下面消成0
        for (int i = r + 1; i <= n; i ++ )
            for (int j = n + 1; j >= c; j -- )
                b[i][j] -= b[i][c] * b[r][j];
    }

    // 转化成对角矩阵
    for (int i = n; i > 1; i -- )//列开始
        for (int j = i - 1; j; j -- )//行
        {
            b[j][n + 1] -= b[i][n + 1] * b[j][i];//
            b[j][i] = 0;
        }
}
  
//结果是每行的最后一个数
    for (int i = 1; i <= n; i ++ ) printf("%.3lf ", b[i][n + 1]);

https://www.acwing.com/activity/content/problem/content/1775/

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <cmath>

using namespace std;

const int N = 15;

int n;
double a[N][N], b[N][N];//a为原先的系数数组 ，b是整理之后的系数 数组

void gauss()
{
    // 转化成上三角矩阵
    for (int r = 1, c = 1; c <= n; c ++, r ++ )
    {
        // 找主元
        int t = r;
        for (int i = r + 1; i <= n; i ++ )
            if (fabs(b[i][c]) > fabs(b[t][c]))
                t = i;

        // 交换
        for (int i = c; i <= n + 1; i ++ ) swap(b[t][i], b[r][i]);
        // 归一化
        for (int i = n + 1; i >= c; i -- ) b[r][i] /= b[r][c];
        // 消
        for (int i = r + 1; i <= n; i ++ )
            for (int j = n + 1; j >= c; j -- )
                b[i][j] -= b[i][c] * b[r][j];
    }

    // 转化成对角矩阵
    for (int i = n; i > 1; i -- )
        for (int j = i - 1; j; j -- )
        {
            b[j][n + 1] -= b[i][n + 1] * b[j][i];
            b[j][i] = 0;
        }
}

int main()
{
    scanf("%d", &n);
    for (int i = 0; i < n + 1; i ++ )
        for (int j = 1; j <= n; j ++ )
            scanf("%lf", &a[i][j]);

    for (int i = 1; i <= n; i ++ )
        for (int j = 1; j <= n; j ++ )
        {
            b[i][j] += 2 * (a[i][j] - a[0][j]);
            b[i][n + 1] += a[i][j] * a[i][j] - a[0][j] * a[0][j];
        }

    gauss();

    for (int i = 1; i <= n; i ++ ) printf("%.3lf ", b[i][n + 1]);

    return 0;
}

异或版

int gauss()
{
    int r, c;
    for (r = 1, c = 1; c <= n; c ++ )
    {
        // 找主元
        int t = r;
        for (int i = r + 1; i <= n; i ++ )
            if (a[i][c])
                t = i;

        if (!a[t][c]) continue;
        // 交换
        for (int i = c; i <= n + 1; i ++ ) swap(a[t][i], a[r][i]);
        // 消
        for (int i = r + 1; i <= n; i ++ )
            for (int j = n + 1; j >= c; j -- )
                a[i][j] ^= a[i][c] & a[r][j];
        r ++ ;
    }

    int res = 1;
    if (r < n + 1)//有解应该是n+1 没有解的话就是无解 有自由元
    {
        for (int i = r; i <= n; i ++ )
        {
            if (a[i][n + 1]) return -1;  // 出现了 0 == !0，无解
            res *= 2;//每个自由元说明多了两种选择
        }
    }

    return res;
}

标签：转化成,int,矩阵,gauss,--,include,高斯消
来源： https://www.cnblogs.com/liang302/p/16683230.html