模板-高斯消元

Posted 2021-03-06 hznumqf

高斯-约旦消元法

int n;

double a[maxn][maxn];

void Gauss() {
    scanf("%d", &n);
    for (int i = 1; i <=n; i++) {
        for (int j = 1; j <=n + 1; j++) {
            scanf("%lf", &a[i][j]);
        }
    }
    for (int i = 1; i <= n; i++) {   //枚举列(项)
        int max = i;
        for (int j = i + 1; j <= n; j++) {
            if (fabs(a[j][i]) > fabs(a[maxn][i])) max = j;
        }
        for (int j = 1; j <= n + 1; j++) {
            swap(a[i][j], a[max][j]);
        }
        if (!a[i][i]) {          //最大值为0说明该列都为0，无解
            printf("No solution"); return;
        }
        for (int j = 1; j <= n; j++) {    //每一项都减去一个数
            if (j != i) {
                double tmp = a[j][i] / a[i][i];
                for (int k = i + 1; k <= n + 1; k++) {
                    a[j][k] -= a[i][k] * tmp;
                }
            }
        }
    }
    for (int i = 1; i <= n; i++) {
        printf("%.2f
", a[i][n + 1] / a[i][i]);
    }
}