answer:Here are some things to be aware of when using `std::unordered_map` in C++: 1. **Hash function**: The performance of `std::unordered_map` relies heavily on the quality of the hash function. A good hash function should distribute keys evenly across the map to minimize collisions. If you're using a custom type as a key, you may need to provide a custom hash function. 2. **Load factor**: The load factor is the ratio of the number of elements to the number of buckets. If the load factor becomes too high, the map may become slow due to increased collisions. You can check the load factor using the `load_factor()` function and rehash the map using the `rehash()` function if necessary. 3. **Resizing**: When the map grows or shrinks, it may need to resize its internal array of buckets. This can be an expensive operation, especially if the map is large. You can minimize the number of resizes by providing an initial size hint when creating the map. 4. **Iterator invalidation**: When the map is modified, iterators may become invalid. This is especially true if the map is resized. You should be cautious when using iterators and check the documentation to see when they may become invalid. 5. **Cache locality**: `std::unordered_map` uses a hash table, which can lead to poor cache locality. This can result in slower performance compared to other data structures like `std::vector` or `std::array`. 6. **Insertion order**: Unlike `std::map`, `std::unordered_map` does not preserve the insertion order of elements. If you need to preserve the insertion order, you may want to consider using `std::map` or another data structure. 7. **Exception safety**: `std::unordered_map` provides the strong exception safety guarantee, which means that if an exception is thrown during an operation, the map will remain in a valid state. However, you should still be cautious when using exceptions with `std::unordered_map` and check the documentation to see how exceptions may affect the map's state. 8. **Custom key types**: When using custom types as keys, you need to provide a custom comparison operator (e.g., `operator==`) and a custom hash function. You should also ensure that the custom type is copyable and movable. 9. **Thread safety**: `std::unordered_map` is not inherently thread-safe. If you need to access the map from multiple threads, you will need to use synchronization mechanisms like locks or atomic operations. 10. **C++ version**: The behavior of `std::unordered_map` has changed slightly between C++11 and C++14. In C++14, the map's `size()` function is required to be constant time, whereas in C++11 it may be linear time in the worst case.

answer:To find the remainder when 99^99 is divided by 11, we can apply modular arithmetic and Fermat's Little Theorem. First, observe that 99 is congruent to -2 modulo 11, meaning 99 ≡ -2 (mod 11). Therefore, we can rewrite 99^99 as (-2)^99. Now, since (-2)^2 = 4, we can further simplify this expression. We note that 99 is an odd number, so (-2)^99 is equal to (-2) * (4)^49. We're now interested in finding the remainder of 4^49 modulo 11. Fermat's Little Theorem tells us that if p is a prime number and a is not divisible by p, then a^(p-1) ≡ 1 (mod p). Applying this theorem to our case with p = 11 and a = 4, we have 4^(11-1) = 4^10 ≡ 1 (mod 11). We can then express 4^49 as (4^10)^4 * 4^9, which simplifies to 1^4 * 4^9 ≡ 4^9 (mod 11). We continue simplifying 4^9 as (4^5) * (4^4), and then apply modular arithmetic again. Since 4^5 = 1024 ≡ 1 (mod 11) and 4^4 = 256 ≡ 3 (mod 11), we obtain 4^9 ≡ 1 * 3 ≡ 3 (mod 11). Combining the results, we have (-2) * 3 ≡ -6 ≡ 5 (mod 11). Hence, the remainder when 99^99 is divided by 11 is 5.

answer:To find the remainder when 2^125 is divided by 13, we can use Fermat's Little Theorem. Since 13 is a prime number and 2 is not divisible by 13, we know that 2^(13-1) ≡ 1 (mod 13). This means 2^12 ≡ 1 (mod 13). We're interested in finding the remainder of 2^125 modulo 13. To utilize Fermat's Little Theorem, we rewrite 125 as 125 = 12 * 10 + 5. Then, 2^125 = (2^12)^10 * 2^5. Since 2^12 ≡ 1 (mod 13), we can simplify this expression as (1)^10 * 2^5 ≡ 2^5 (mod 13). Calculating 2^5 = 32, we note that 32 ≡ 6 (mod 13) because 32 - 26 = 6. Hence, the remainder when 2^125 is divided by 13 is 6 modulo 13, which simply means the remainder is 6.

answer:To find the remainder when 7^123 is divided by 11, we can use Fermat's Little Theorem. Since 11 is a prime number and 7 is not divisible by 11, we know that 7^(11-1) ≡ 1 (mod 11). This means 7^10 ≡ 1 (mod 11). We're interested in finding the remainder of 7^123 modulo 11. To utilize Fermat's Little Theorem, we rewrite 123 as 123 = 12 * 10 + 3. Then, 7^123 = (7^10)^12 * 7^3. Since 7^10 ≡ 1 (mod 11), we can simplify this expression as (1)^12 * 7^3 ≡ 7^3 (mod 11). Calculating 7^3 = 343, we note that 343 ≡ 2 (mod 11) because 343 - 341 = 2. Hence, the remainder when 7^123 is divided by 11 is 2.