#include<stdio.h>
int main()
{
    int i = 0, j = 0, n = 0, m;
    char a[80];
    int b[80];
    gets_s(a);
    while (a[i] != '\0')
    {
        if (a[i] >= '0'&&a[i] <= '9')
        {
            m = 10 * n + (a[i] - '0');
            n = m;
        }
        if ((a[i] >= '0'&&a[i] <= '9') && (a[i + 1] < '0'&&a[i + 1] > '9'))
        {
            b[j] = m;
            j++;
            m = 0;
        }
        i++;
    }
    for (i = 0; i < j; i++)
        printf("%d ", b[i]);
    printf("\n");
    return 0;
}

做出来了，嘎嘎

#include<stdio.h>
int main()
{
    int i, j = 0, m = 0;
    char a[80];
    int b[80];
    gets_s(a);
    for (i = 0; a[i] != '\0'; i++)
    {
        if (a[i] >= '0'&&a[i] <= '9')
            m = 10 * m + (a[i] - '0');
        if ((a[i] >= '0'&&a[i] <= '9') && (a[i + 1] < '0'||a[i + 1] > '9'))
        {
            b[j] = m;
            j++;
            m = 0;
        }
    }
    for (i = 0; i < j; i++)
        printf("%d ", b[i]);
    printf("\n");
    printf("有%d个整数\n", j);
    return 0;
}

修改版，答案还没看懂

#include<stdio.h>
void input(char *p1)
{
    gets_s(p1, 80);
}

void execute(char *p1, int *p2)
{
    int i = 0, j = 0, m = 0;
    while (p1[i++] != '\0')
    {
        if (p1[i] >= '0'&&p1[i] <= '9')
            m = 10 * m + (p1[i] - '0');
        if ((p1[i] >= '0'&&p1[i] <= '9') && (p1[i + 1]<'0' || p1[i + 1]>'9'))
        {
            p2[j++] = m;
            m = 0;
        }
    }
    for (i = 0; i < j; i++)
        printf("%d%c", p2[i], i < j - 1 ? ' ' : '\n');
    printf("There are %d numbers in this line\n", j);
}

int main()
{
    void input(char *p1);
    void execute(char *p1, int *p2);
    char a[80], *p1;
    int b[80], *p2;
    p1 = a;
    p2 = b;
    input(p1);
    execute(p1, p2);
    return 0;
}

printf("%d%c", p2[i], i < j - 1 ? ' ' : '\n');

原来答案也会开挂~连个换行符输出都写在主要的输出语句中~

感觉指针运用得较多~

感觉用%c输出数字会比较方便一些~