Question

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <ctype.h>
#define N 10
#define TEMPSIZE 1024

int DiagSum(int **p, int sizearray) {
    int sum=0;
    for (int i = 0; i < sizearray; i++) {
        for (int j = 0; j < sizearray; j++) {
            if (j == i) {
                sum = sum + p[i][j];
            }
        }
    }
    return sum;
}

int SizeArray(int **p) {
    int i;
    for (i = 0; i < TEMPSIZE; i++) {
        if (p[i] == ' ' || p[i]=='\t')
            i++;
    }
    return p[i];
}
void main() {
    int **p;
    int input;
    int sizear=0;
    int i, j, counter = 0;

    p = (int**)malloc(N * sizeof(int*));
    for (j = 0; j < N; j++)
        p[j] = (int*)malloc(N * sizeof(int));
    while ((input = getchar()) != EOF || input != '\n') {
        if (input == ' ' || input == '\t')
            continue;
        input = input - '0';
        sizear = SizeArray(p);
        for (i = 0; i < sizear; i++)
        {
            for (j = 0; j < sizear; j++)
            {
                p[i][j] = input - '0';
                counter++;
            }
        }
    if (counter > sizear*sizear) {
        printf("you've entered too many numbers \n");
    }
    else if (counter < sizear*sizear) {
        printf("you've entered not enough numbers \n");
    }
    for (i = 0; i < sizear; i++)
    {
        for (j = 0; j < sizear; j++)
        {
            printf("%c ", p[i][j]);
        }
    }
}

完全陷入困境。我的作业要求我从用户那里获得输入，只有一行。它将由矩阵的大小（n * n）组成，它将是该行中的第一个数字，以及它中的数字。例如：3 1 2 3 4 5 6 7 8 9. 3（第一个数字） - 列和列，后面的数字将是矩阵内的数字。第一个错误，最重要的是无法提取第一个数字，即数组大小，第二个可能是输入到矩阵中的错误。我该如何解决？

Answer 1

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <ctype.h>
#define TEMPSIZE 1024

int DiagSum(int **p, int sizearray) {
    int sum=0;
    for (int i = 0; i < sizearray; i++) {
        for (int j = 0; j < sizearray; j++) {
            if (j == i) {
                sum = sum + p[i][j];
            }
        }
    }
    return sum;
}

void main() {
    int **p;
    char input[TEMPSIZE] = {0};
    int sizear=0;
    int i, j, counter = 0;

    //get input string(size no more than 1024)
    fgets(input, TEMPSIZE, stdin);

    //get array size
    char* str = strstr(input, " ");
    if(str == 0)
    return;
    str[0] = 0;
    str++;
    sizear = atoi(input);

    if(sizear <= 0)
    return;

    //create array
    p = (int**)calloc(sizear, sizeof(int*));
    for (j = 0; j < sizear; j++)
    {
        p[j] = (int*)calloc(sizear, sizeof(int));
    }

    //fill array with input
    for (i = 0; i < sizear; i++)
    {
        for (j = 0; j < sizear; j++)
        {
            char* temp = strstr(str, " ");
            if(temp == 0)
            {
                p[i][j] = atoi(str);
                counter++;
                goto end;
            }
            temp[0] = 0;
            p[i][j] = atoi(str);
            str = temp + 1;
            counter++;
        }
    }

end:
    if (counter > sizear*sizear) {
        printf("you've entered too many numbers \n");
    }
    else if (counter < sizear*sizear) {
        printf("you've entered not enough numbers \n");
    }
    for (i = 0; i < sizear; i++)
    {
        for (j = 0; j < sizear; j++)
        {
            printf("%d ", p[i][j]);
        }
    }

    //free memory
    for (j = 0; j < sizear; j++)
    {
        free(p[j]);
    } 
    free(p);
}

我认为上面的代码可以帮到你！

从二维数组中获取用户输入

1 个答案: