杜利特尔分解Doolittle转化为克洛特分解Crout_解线性方程组的直接解法

杜利特尔分解Doolittle转化为克洛特分解Crout_解线性方程组的直接解法

标签：计算方法实验

#include 
#include 
#include 

const int maxn = 15;

int main(){
    double a[maxn][maxn], b[maxn], y[maxn], x[maxn], l[maxn][maxn], u[maxn][maxn], d[maxn];
    int i, j, k, r, n, sum;

    freopen("lu.txt", "r", stdin);
    scanf("%d", &n);
    for(int i = 1; i <= n; i++){
        for(int j = 1; j <= n; j++)  scanf("%lf", &a[i][j]);
        scanf("%lf", &b[i]);
    }

    for(i = 1; i <= n; i++)  u[1][i] = a[1][i];  //U的第一行元素
    for(i = 1; i <= n; i++)  l[i][1] = a[i][1] / u[1][1];  //L的第一列元素

    for(k = 2; k <= n; k++){
        for(j = k; j <= n; j++){  //计算U的第k行元素
            for(r = 1, sum = 0; r <= k - 1; r++)  sum += (l[k][r] * u[r][j]);
            u[k][j] = a[k][j] - sum;
        }

        for(i = k; i <= n; i++){  //计算L的第k列元素
            for(r = 1, sum = 0; r <= k - 1; r++)  sum += (l[i][r] * u[r][k]);
            l[i][k] = (a[i][k] - sum) / u[k][k];
        }
    }

    for(i = 1; i <= n; i++){  //打印L   U
        for(j = 1; j <= n; j++)  printf("%10f", l[i][j]);
        printf("\t\t");
        for(k = 1; k <= n; k++)  printf("%10f", u[i][k]);
        printf("\n");
    }

    /*
    y[1] = b[1];  //求解Ly = b
    for(i = 2; i <= n; i++){
        for(k = 1, sum = 0; k <= i - 1; k++)  sum += (l[i][k] * y[k]);
        y[i] = b[i] - sum;
    }

    x[n] = y[n] / u[n][n];  //求解Ux = y
    for(i = n - 1; i >= 1; i--){
        for(k = i + 1, sum = 0; k <= n; k++)  sum += (u[i][k] * x[k]);
        x[i] = (y[i] - sum) / u[i][i];
    }

    for(int i = 1; i <= n; i++)  printf("%10f\n", x[i]);
    */

    //A = LDD^-1U(杜利特尔) -> A = (LD)(D^-1U)(克洛特)
    //d存储u对角线上的元素
    for(i = 1; i <= n; i++)  d[i] = u[i][i];

    for(i = 1; i <= n; i++){  //L = (LD)
        for(j = 1; j <= i; j++)  l[i][j] *= d[j];
    }
    for(i = 1; i <= n; i++){  //U = (D^-1U)
        for(j = i; j <= n; j++)  u[i][j] /= d[i];  //d的逆d^-1[i] = 1 / d[i]
    }

    printf("杜利特尔分解(Doolittle)->克洛特分解(Crout):\n");
    for(i = 1; i <= n; i++){  //打印L   U
        for(j = 1; j <= n; j++)  printf("%10f", l[i][j]);
        printf("\t\t");
        for(k = 1; k <= n; k++)  printf("%10f", u[i][k]);
        printf("\n");
    }

    /*
    memset(y, 0, sizeof(y)), memset(x, 0, sizeof(x));
    y[1] = b[1] / l[1][1];  //求解Ly = b
    for(i = 2; i <= n; i++){
        for(k = 1, sum = 0; k <= i; k++)  sum += (l[i][k] * y[k]);
        y[i] = (b[i] - sum) / l[i][i];
    }

    x[n] = y[n];  //求解Ux = y
    for(i = n - 1; i >= 1; i--){
        for(k = i + 1, sum = 0; k <= n; k++)  sum += (u[i][k] * x[k]);
        x[i] = y[i] - sum;
    }

    for(int i = 1; i <= n; i++)  printf("%10f\n", x[i]);
    */

    return 0;
}

数据文件

实验结果

去掉/* */注释后