克洛特分解法Crout_解线性方程组的直接解法

标签：计算方法实验

#include 
   
    #include 
    
     const int maxn = 15;int main(){    double a[maxn][maxn], b[maxn], y[maxn], x[maxn], l[maxn][maxn], u[maxn][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]);    }    /*打印数据文件    for(int i = 1; i <= n; i++){        for(int j = 1; j <= n; j++)  printf("%10f", a[i][j]);        printf("%10f\n", b[i]);    }    */    for(i = 1; i <= n; i++)  l[i][1] = a[i][1];  //L的第一列元素    for(i = 1; i <= n; i++)  u[1][i] = a[1][i] / l[1][1];  //U的第一行元素    for(k = 2; k <= n; k++){        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;        }        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) / l[k][k];        }    }    /*打印L   U    for(i = 1; i <= n; i++){        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] / 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;}

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