Exponentiation by squaring still “works” for modulo exponentiation. Your problem isn’t that 2 ^ 168277 is an exceptionally large number, it’s that one of your intermediate results is a fairly large number (bigger than 2^32), because 673109 is bigger than 2^16. So I think the following will do. It’s possible I’ve missed a detail, but … Read more
In MATLAB, one option you have is to overload the methods that handle arithmetic operations for integer data types, creating your own custom overflow behavior that will result in a “wrap-around” of the integer value. As stated in the documentation: You can define or overload your own methods for int* (as you can for any … Read more
To check for over/underflow in arithmetic check the result compared to the original values. uint32 a,b; //assign values uint32 result = a + b; if (result < a) { //Overflow } For your specific the check would be: if (a > (c-b)) { //Underflow }
Think of this function: int f(int i) { return i+1 > i; } Mathematically speaking, i+1 should always be greater than i for any integer i. However, for a 32-bit int, there is one value of i that makes that statement false, which is 2147483647 (i.e. 0x7FFFFFFF, i.e. INT_MAX). Adding one to that number will … Read more
I’m going to assume that the online compilers use GCC or compatible compiler. Of course, any other compiler is also allowed to do the same optimization, but GCC documentation explains well what it does: -faggressive-loop-optimizations This option tells the loop optimizer to use language constraints to derive bounds for the number of iterations of a … Read more
If you need larger than 32-bits you could consider using 64-bit integers (long long), or use or write an arbitrary precision math library, e.g. GNU MP.
Signed overflow is undefined in C, and that’s for real. One solution follows: signed_result = (unsigned int)one_argument + (unsigned int)other_argument; The above solution involves implementation-defined behavior in the final conversion from unsigned to int but do not invoke undefined behavior. With most compilation platforms’ implementation-defined choices, the result is exactly the two’s complement result that … Read more
This image shows what you’re looking for. In your case it’s obviously larger numbers, but the principle stays the same. Examples of limits in java are: int: −2,147,483,648 to 2,147,483,647. long: -9,223,372,036,854,775,808 to 9,223,372,036,854,775,807 In the image 0000, 0001 etc, shows the binary representation of the numbers. EDIT: In project euler you often have to … Read more
Signed integer overflow is undefined behaviour, while unsigned integer overflow is well-defined; the value wraps around. In other words, the value is modulo divided by 2bits, where bits is the number of bits in the data type. Since you’ve a 32-bit int 4294967295 + 1 = 4294967296 % 232 = 0 it results in 0 … Read more
unsigned numbers can’t overflow, but instead wrap around using the properties of modulo. For instance, when unsigned int is 32 bits, the result would be: (a * b) mod 2^32. As CharlesBailey pointed out, 253473829*13482018273 may use signed multiplication before being converted, and so you should be explicit about unsigned before the multiplication: unsigned int … Read more