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`You are batting at the pitch and all eyes are on you, the best player in all of IPL The opposing team will relentlessly bowl at you an infinite number of balls You notice that you only ever score a boundary on the Mth ball, if N mod M?N/2, where N is the lucky number of your T-shirt. Now you wanna know how many boundaries will you hit in total? Formally, you are given a positive integer N and you need to find the number of positive integers M such that N mod M?N/2.`

What I have tried:

```#include <iostream>
Using namespace std;
int main()
{
int T:
C in>>T;
for(int j =0;j<T; j++)
{
int N:
C in>>N;
```
Posted
Updated 17-Jul-22 0:38am
CHill60 13-Jul-22 4:51am
Richard MacCutchan 13-Jul-22 4:52am
What exactly is "N mod M?N/2"? I know `N mod M` and `N / 2`, but what does the `?` mean in that expression?
Patrice T 13-Jul-22 4:56am
And you probably have a question and context too.

Solution 1

Im not an expert here I dont understand the question either but
this is my understanding of the statement above..

this will display the numbers from 0 to Mth value where N is the lucky number and M is the number od iterations
then the M modulo N should be equal to the value of (N/2)

the program below will result to
```the M = 7
the M = 22
the M = 37
the M = 52
the M = 67
the M = 82
the M = 97
Number of Hits: 7```

=============================================================
```#include <stdio.h>

int M_limit=100;    //the Mth limit
int N = 15;         //your lucky number N
int H_count=0;      //number positive integers M

void the_M(){

for(int a=1; a<M_limit; a++){

int M=a;
if( (N%M) ==(N/2)){             //if we reverse M and N in the modulo the program will crash i dont know
printf("the M = %d\n",a);   //print the M value based from the formula above i think
H_count++;                  //
}
}
printf("\nNumber of Hits: %d\n",H_count);  //print the number of integers of Ms
}

int main(){

the_M();
return 0;
}```

v2
CPallini 13-Jul-22 9:09am
"//if we reverse M and N in the modulo the program will crash i dont know"
Because division by 0 is not allowed (as in ordinary math...).
Note that Requirements state: "you need to find the number of positive integers M", so, starting with a=0 doesn't make sense.
bonezegei 13-Jul-22 9:17am
oh, yeah right. I just realize that now. Thank you

therefore it should be like this,
I tested it just now and it worked..

for(int a=1; a
CPallini 13-Jul-22 9:18am
You are welcome.
Dave Kreskowiak 13-Jul-22 9:45am
Doing someones homework for them is not helping them and is never a good idea.
bonezegei 13-Jul-22 10:00am
yes that's really true. But on the otherside I learned from it.

Solution 2

As this is obviously a far - fetched story and formula based on it, one cannot be sure what the expected solution will be. Maybe the values ​​don't matter at all...
Based on the source code given by the author and the solution formula (N%M) ? (N/2) a solution in C++ could look like this.
C++
```cin >> N;
int M_MAX = N * 2; // limit of M
for (int M = 1; M <= M_MAX; M++) {
if( (N%M) ? (N/2) : 0 > 0) {
cout << M << ", ";    // print the M value if it fits
count++;              // count boundaries
}
}```

// Output for Lucky Number N = 15 would be 2, 4, 6, 7, 8, 9, 10, ...
Number of Hits : 26

v2