Question

char input1[1000], input2[1000];

while(scanf("%s %s", input1, input2)!=EOF) 
{
    //input1 = -3x^2+5x-4
    //input2 = x^2-1
}

在此之上，我想输入一个如下所示的多项式

// Create first list of -3x^2+5x-4
create_node(-3,2,&poly1);
create_node(5,1,&poly1);
create_node(-4,0,&poly1);
// Create second list of x^2-1
create_node(1,2,&poly2);
create_node(-1,0,&poly2);

我的问题是：（-3x ^ 2 + 5x-4）例如

我如何检测到-3、2、5，-4

create_node(-3,2,&poly1);
create_node(5,1,&poly1);
create_node(-4,0,&poly1);

我的想法是如何检测

//input 1 = -3x^2+5x-4

直到x的前面（将是-3），并检测到^的后面（将是2），依此类推。

但是我如何检测字符x和^等。

这是我的完整代码my code

谢谢

Answer 1

#include <ctype.h>
#include <stdlib.h>
#include <stdio.h>
#include <errno.h>

int main(void)
{
    char buffer[1000];
    while(fgets(buffer, 1000, stdin)) {

        char *p = buffer;
        while (*p) {
            while (*p && isspace(*p))
                ++p;

            if (!*p)
                break;

            if (!isdigit(*p)) {
                printf("Variable/operator: %c\n", *p++);
            }
            else {
                char *end;
                errno = 0;
                long number = strtol(p, &end, 10);

                if (errno == ERANGE) {
                    fputs("Value out of range!\n", stderr);
                    break;
                }
                else if(p != end) {
                    printf("Value: %ld\n", number);
                    p = end;
                }
            }
        }
    }
}

输出：

-3x^2+5x-4
Variable/operator: -
Value: 3
Variable/operator: x
Variable/operator: ^
Value: 2
Variable/operator: +
Value: 5
Variable/operator: x
Variable/operator: -
Value: 4

Answer 2

在create_node 28行的功能中，r未初始化
为什么要在列表末尾添加空节点？

以下代码可以解决您的问题和两个问题。

#include <stdio.h>
#include <stdlib.h>

struct Node
{
    int coeff;
    int pow;
    struct Node *next;
};

// Function to create new node
void create_node(int x, int y, struct Node **temp)
{
    struct Node *r = *temp;
    if(r == NULL)
    {
        r =(struct Node*)malloc(sizeof(struct Node));
        r->coeff = x;
        r->pow = y;
        r->next = NULL;
        *temp = r;
    }
    else
    {
        while (r->next != NULL)
            r = r->next;
        r = r->next = (struct Node*)malloc(sizeof(struct Node));
        r->coeff = x;
        r->pow = y;
        r->next = NULL;
    }
}

// Function Adding two polynomial numbers
void polyadd(struct Node *poly1, struct Node *poly2, struct Node **poly3)
{
    struct Node* poly;
    while(poly1 && poly2)
    {
        // Dynamically create new node
        if (*poly3 == NULL) {
            poly = (struct Node *)malloc(sizeof(struct Node));
            *poly3 = poly;
        } else {
            poly = poly->next = (struct Node *)malloc(sizeof(struct Node));
        }
        poly->next = NULL;

        // If power of 1st polynomial is greater then 2nd, then store 1st as it is
        // and move its pointer
        if(poly1->pow > poly2->pow)
        {
            poly->pow = poly1->pow;
            poly->coeff = poly1->coeff;
            poly1 = poly1->next;
        }

        // If power of 2nd polynomial is greater then 1st, then store 2nd as it is
        // and move its pointer
        else if(poly1->pow < poly2->pow)
        {
            poly->pow = poly2->pow;
            poly->coeff = poly2->coeff;
            poly2 = poly2->next;
        }

        // If power of both polynomial numbers is same then add their coefficients
        else
        {
            poly->pow = poly1->pow;
            poly->coeff = poly1->coeff+poly2->coeff;
            poly1 = poly1->next;
            poly2 = poly2->next;
        }

    }
    while(poly1 || poly2)
    {
        if (*poly3 == NULL) {
            poly = (struct Node *)malloc(sizeof(struct Node));
            *poly3 = poly;
        } else {
            poly = poly->next = (struct Node *)malloc(sizeof(struct Node));
        }
        poly->next = NULL;
        if(poly1)
        {
            poly->pow = poly1->pow;
            poly->coeff = poly1->coeff;
            poly1 = poly1->next;
        }
        if(poly2)
        {
            poly->pow = poly2->pow;
            poly->coeff = poly2->coeff;
            poly2 = poly2->next;
        }
    }
}

void polySub(struct Node *poly1, struct Node *poly2, struct Node *poly)
{

}

// Display Linked list
void show(struct Node *node)
{
    while(node != NULL)
    {
        if (node->coeff != 1 && node->pow != 0) {
            printf("%dx^%d", node->coeff, node->pow);
        } else if (node->coeff == 1 && node->pow != 0) {
            printf("x^%d", node->pow);
        } else if (node->coeff != 1 && node->pow == 0) {
            printf("%d", node->coeff);
        }
        node = node->next;
        if(node != NULL)
            printf(" + ");
    }
}

void detext_string(char *st, struct Node **poly) {
    int coeff;
    int pow;

    while (*st != '\0') {
        coeff = strtol(st, &st, 10);
        if (coeff == 0)
            coeff = 1;

        if (*st == 'x') {
            ++st;
            if (*st == '^') {
                ++st;
                pow = strtol(st, &st, 10);
            } else {
                pow = 1;
            }
        } else {
            pow = 0;
        }
        create_node(coeff, pow, poly);
    }
}

void free_node(struct Node** p) {
    struct Node* q = *p;
    while (q != NULL) {
        struct Node* temp = q;
        q = q->next;
        free(temp);
    }
    *p = NULL;
}

// Driver  program
int main()
{
    struct Node *poly1 = NULL, *poly2 = NULL, *poly = NULL;

    char input1[1000], input2[1000];

    while(scanf("%s %s", input1, input2) == 2)
    {
        detext_string(input1, &poly1);
        detext_string(input2, &poly2);

        printf("1st Number: ");
        show(poly1);

        printf("\n2nd Number: ");
        show(poly2);

        // Function add two polynomial numbers
        polyadd(poly1, poly2, &poly);

        // Display resultant List
        printf("\nAdded polynomial: ");
        show(poly);
        printf("\n");
        free_node(&poly1);
        free_node(&poly2);
        free_node(&poly);
    }
    return 0;
}

检测c中的字符

2 个答案:

输出：