本文共 2627 字，大约阅读时间需要 8 分钟。

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