NOJ1004线性表操作
解题思路暴力，来多少放多少。
(下列代码已经通过测试)
#include
#include
int main()
{
    int che1, che2, che3;
    int* in = NULL;
    int dein = 0;
    char* cha = NULL;
    char decha = 0;
    double* dou = NULL;
    double dedou = 0;
    int n1, n2, n3;
    {
        scanf("%d\n", &n1);
        in = malloc(sizeof(int) * n1);
        for (che1 = 0; che1 < n1; che1++)
            scanf("%d", in + che1);
        scanf("%d", &dein);
    }
    {
        scanf("%d\n", &n2);
        cha = malloc(sizeof(char) * n2);
        for (che2 = 0; che2 < n2; che2++)
            scanf("%c ", cha + che2);
        scanf("%c", &decha);
    }
    {
        scanf("%d\n", &n3);

        dou = malloc(sizeof(double) * n3);
        for (che3 = 0; che3 < n3; che3++) scanf("%lf", dou + che3); scanf("%lf", &dedou); } //int { for (che1 = n1 - 1; che1 >= 0; che1--)
            printf("%d ", in[che1]);
        printf("\n");
        for (che1 = n1 - 1; che1 >= 0; che1--)
        {
            if (in[che1] == dein)
                continue;
            printf("%d ", in[che1]);
        }
        printf("\n");
    }
    //char
    {
        for (che2 = n2 - 1; che2 >= 0; che2--)
            printf("%c ", cha[che2]);
        printf("\n");
        for (che2 = n2 - 1; che2 >= 0; che2--)
        {
            if (cha[che2] == decha)
                continue;
            printf("%c ", cha[che2]);
        }
        printf("\n");
    }
    //double
    {        for (che3 = n3 - 1; che3 >= 0; che3--)
        printf("%g ", dou[che3]);
    printf("\n");
    for (che3 = n3 - 1; che3 >= 0; che3--)
    {
        if (dou[che3] - dedou < 1E-6)
            continue;
        printf("%g ", dou[che3]);
    }
 }
    return 0;
}