int overflows

Topics

• c++ internals

Add some value to int or long beyond their max capacity, results in the excess offset from the minimum value.

INT_MAX + N = (INT_MAX + 1) + N - 1
INT_MAX + N = (-INT_MAX - 1) + (N - 1) = -INT_MAX + N - 2

In above example, if we have N=INT_MAX, we get:

int x = INT_MAX;
int N = INT_MAX;
cout << (x+N) << endl; // -2

In many coding contests, a nice way to deal with overflows is to use modular arithmetic to keep value within intlimit (1e9 + 7 is a popular choice to mod with).

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