**Write a Modular C Programming code to find out the efficiency of the worker**

In a company, worker efficiency is determined on the basis of the time required for a worker to complete a particular job. If the time taken by the worker is between 2-3 hours, then the worker is said to be highly efficient. If the time required by the worker is between 3-4 hours, then the worker is ordered to improve speed. If the time taken is between 4-5 hours, the worker is given the training to improve his speed, and if the time taken by the worker is more than 5 hours, then the worker is terminated.

The time taken by the worker is input through the keyboard; write a c program to find the efficiency of the worker.

**Input Format**

time to complete a job in hours and minutes (hh mm).

**Constraints**

0<=hh<=24 0<=mm<=59

**Output Format**

print work efficiency of worker

**Sample Input 0**

```
2 00
```

**Sample Output 0**

```
highly efficient
```

**Sample Input 1**

```
3 40
```

**Sample Output 1**

```
improve speed
```

**Sample Input 2**

```
4 45
```

**Sample Output 2**

```
training to improve speed
```

**Sample Input 3**

```
5 15
```

**Sample Output 3**

```
terminated
```

**Sample Input 4**

```
25 25
```

**Sample Output 4**

`Invalid time`

**Refer :** C Programming HackerRank all solutions for Loops | Arrays | strings

**CODE:**

#include<stdio.h> #include<stdlib.h> #include<string.h> #include<math.h> int main() { int h,m; scanf("%d%d",&h,&m); if(h>=0&&h<=24&&m>=0&&m<=59) { if(h>=2&&h<3&&m>=0&&m<=59) printf("highly efficient"); else if(h>=3&&h<4&&m>=0&&m<=59) printf("improve speed"); else if(h>=4&&h<5&&m>=0&&m<=59) printf("training to improve speed"); else printf("terminated"); } else printf("Invalid time"); }

**OUTPUT**

Congratulations, you passed the sample test case. Click the Submit Code button to run your code against all the test cases. Input (stdin) 2 00 Your Output (stdout) highly efficient Expected Output highly efficient

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