Thread: help w/function

for (t = 0; t < 10; ++t)
{
for (i = 0; i < 10; ++i)


since you take in the value for size, you might as well use it here for 10.

Problem now, it keeps looping.

realize that it is going to say something ~100 times with way you have it set up.

pull the "sorry...." outside of the loop like this

Code:

for (t = 0; t < 10; ++t) 
{ 
   for (i = 0; i < 10; ++i) 
   { 
      if (array[t] + array[i] == 100) 
      {
         cout << "The numbers equal 100.\n";
         set = 1;
       } 
    }
    if (set==1)
      cout << "Sorry, I cannot find 2 numbers whose sum is 100.\n"; 

}

or even better

Code:

int set=0;

for (t = 0; t < 10; ++t) 
{ 
   for (i = 0; i < 10; ++i) 
   { 
      if (array[t] + array[i] == 100) 
      {
         cout << "The numbers equal 100.\n";
         set++;
       } 
    }
    if (set==0)
      cout << "Sorry, I cannot find 2 numbers whose sum is 100.\n"; 
    else
       cout << "Found " << set << " sets of numbers whose sum is 100\n."

}

Once again this doesn't take into account one number equaling 50. Notice how one number equaling 50 will make it appear that 100 was reached using two different numbers.

I will leave you to the rest....

A final hint on the problem of calculating the value of the same number if it equals 50...

if/then hint: if number 1 is equal to number 2 then it is the same array position... or rather if number 1 is not equal to number 2 then perform desired operation