列印在範圍0-99之間缺少的元素

它將顯示使用者輸入的給定集中缺少的值

Given : array = {88, 105, 3, 2, 200, 0, 10};
Output : 1 4-9 11-87 89-99

演算法

START
STEP 1-> Take an array with elements, bool flag[MAX] to Fale, int i, j, n to size of array
Step 2-> Loop For from I to 0 and i<n and i++
   IF array[i] < 100 && array[i]>=0
      Set flag[array[i]]=true
   End IF
Step 3 -> End For Loop
Step 4 -> Loop For from i to 0 and i<MAX and ++i
   IF flag[i] == false
      Set j=i+1
      Loop While j<MAX && flag[j] == false
         Set j++
      End While
      If j=i+1
         Print i
      End IF
   Else
      Print i and j-1
   End Else
   Set i=j
End IF
Step 5 -> End For Loop
STOP

範例

#include <stdio.h>
#define MAX 100
int main(int argc, char const *argv[]) {
   int array[] = {88, 105, 3, 2, 200, 0, 10};
   bool flag[MAX] = { false }; //Initializing all the values of flag as false
   int i, j, n;
   n = sizeof(array)/sizeof(array[0]);
   for (i = 0; i < n; i++) {
      if (array[i] < 100 && array[i]>=0) {
         flag[array[i]] = true; //Making the value of the elements present in an array as true, So missing will remain false
      }
   }
   for (i = 0; i < MAX; ++i) {
      if(flag[i] == false) { //Checking for false values
         j = i+1; //Giving the value of the next iteration
         while(j<MAX && flag[j] == false) //Checking the value of flag[j] is false
         j++;
         if (j==i+1) //For printing the missing number
            printf("%d</p><p>", i);
         else //For printing the missing range
            printf("%d-%d</p><p>", i, j-1);
         i = j; //Initializing the range&#39;s last value to start from that number
      }
   }
   return 0;
}

輸出

如果我們執行上面的程序，它將產生以下輸出

1
4-9
11-87
89-99

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