CareerCup

I tried:

Round(7 * (rand()/rand()))

7 * (rand()/rand()) is not right.
There's someone have posted, as far as I know, you may make two succesive call of rand()to produce a number. so you get a random number from 1 to 25, the range is larger than [1,7]. if the number come out is within [8,25] you can simply ignore it.
int getRand() {
int num = 0;
do {
num = (rand()-1)*5+rand();
}while(num > 7);
return num;
}
of course, you are recommented make 5, 7 as parameter of the fuction #getRand().

May you good luck!
/Dam775

How about - rand() + rand()%3 (% is modulo operation)?

rand() + ( (rand()-1)/2 ) , this will work i think.

another combination could be:
let the function is
Rand1to5() // it returns random number between 1 and 5;
then,
Rand1to7 = Rand1to5()+ceil(Rand1to5 * 3 / 5) - 1; //this will always return desired random number

how about dividing the result by 5 and multiplying by 7

None of the ideas above would work because implicit in the question is the fact that the output should be uniformly distributed.

But using rand5, you can get a 0 or a 1 uniformly dist. (if (rand5=1 or 2) output 0, if(rand5=3 or 4) output 1, else try again). calling this 3 times you can get any number from 0 to 7 (in binary). Just ignore the 0 (i.e. generate again)

This would do it. Not the most elegant solution, but WTF.

float a = Rand(); //uniform(1-5)
float b = Rand()/5; //uniform(0-1)
float c = Rand()/5; //unoform(0-1)
if (c>= 1/2) a = a+b;
else a = a-b;