The 50-power algorithm of C ++ 2

I did not know how to solve this problem when I participated in the ACM competition last year. I finally solved it today. It was so simple that I only needed _ int64

 

# Include <iostream>

Using STD: cout;
Using STD: Endl;
Int main ()
{
_ Int64 sum = 0, multiply = 1;

For (INT I = 1; I! = 50; ++ I)
{
Multiply * = 2;
Sum + = multiply;
}
Cout <sum <Endl;
Return 0;

}

_ Int64

 

This is explained in msdn.

_ Int64 nhuge; // declares 64-bit integer

Declare a 64-bit integer, that is, its range is between +-(63 power-1 of 2), so the problem is solved naturally.

 

The running result is as follows: