Question

问题：

第一次扫描是用户请求的转换次数。以下是将要转换的内容。重点是计算编写这些数字所需的LED数量，如下图Numbers in LED

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

int main(){
   int i, j, k, leds, a, clean;

//here i scan the number of times the program will be executed

   scanf ("%i", &a);

   int n[a], l[a];
   char c[101] = {0}; //the string where the numbers will be armazened

   for (i=0; i< a; i++)
   {
       scanf ("%i", &n[i]); //here i get the numbers
   }

   for (i=0; i<a; i++)
   {
       sprintf(c, "%d", n[i]);
       leds = 0; //where the numbers of leds needed will be armazened

       while(c[j] != '\0')
            {
                if(c[j] == '0')
                {
                    leds = leds + 6;
                }
                else if(c[j] == '1')
                {
                    leds = leds + 2;
                }
                else if(c[j] == '2')
                {
                    leds = leds + 5;
                }
                else if(c[j] == '3')
                {
                    leds = leds + 5;
                }
                else if(c[j] == '4')
                {
                    leds = leds + 4;
                }
                else if(c[j] == '5')
                {
                    leds = leds + 5;
                }
                else if(c[j] == '6')
                {
                    leds = leds + 6;
                }
                else if(c[j] == '7')
                {
                    leds = leds + 3;
                }
                else if(c[j] == '8')
                {
                    leds = leds + 7;
                }
                else if(c[j] == '9')
                {
                   leds = leds + 6;
                }
            clean = j;
            j++;

            }


        l[i] = leds; //where the number of leds of each will be armazened to be printed afterwards
   }

   for (i = 0; i<a; i++)
   {
       printf("%i\n", l[i]);
   }

}

我的意见： 3 115380 2819311 23456

我得到的输出： 27（唯一正确的） 2 0

我应该得到什么： 27 29 25

我只是不知道我做错了什么，但我认为它是c []和sprintf。但我该怎么办？

很抱歉，如果英语不对，

Answer 1

您尚未初始化j

   sprintf(c, "%d", n[i]);
   leds = 0; //where the numbers of leds needed will be armazened

应该是

   sprintf(c, "%d", n[i]);
   j = 0;
   leds = 0; //where the numbers of leds needed will be armazened

Answer 2

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

   int main(){
   int i, j, k, leds, a, clean;

   //here i scan the number of times the program will be executed

   scanf ("%i", &a);

   int n[a], l[a];
   char c[101] = {0}; //the string where the numbers will be armazened
   int numLeds[] = {6, 2, 5, 5, 4, 5, 6, 3, 7, 6};

   for (i=0; i< a; i++)
   {
       scanf ("%i", &n[i]); //here i get the numbers
   }

   for (i=0; i<a; i++)
   {
       sprintf(c, "%d", n[i]);
       leds = 0; //where the numbers of leds needed will be armazened

       int j = 0;

       while(c[j] != '\0')
       {
           // 48 = zero in ASCII code
           if (c[j] < 48 || c[j] > 48 + 9)
               continue; 
           int digit = c[j] - 48;
           leds += numLeds[digit];
           clean = j;
           j++;

       }


        l[i] = leds; //where the number of leds of each will be armazened to be printed afterwards
   }

   for (i = 0; i<a; i++)
   {
       printf("%i\n", l[i]);
   }

}

代码不是读取整个字符串

2 个答案: