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1. Predict the output of this program. ``

cpp
#include <cstdint>
#include <iostream>

std::uint64_t fib_two_vars(int n) {
    std::uint64_t prev = 0;  // F(0)
    std::uint64_t curr = 1;  // F(1)
    for (int i = 0; i < n; ++i) {
        std::uint64_t next = prev + curr;
        prev = curr;
        curr = next;
    }
    return prev;
}

int main() {
    std::cout << fib_two_vars(7) << ' ' << fib_two_vars(0) << ' ' << fib_two_vars(2) << '\n';
}

``

13 1 113 0 18 0 113 0 2

2. A string of digits has a 0 at the start, as in "06". Why does "06" have 0 ways to decode?

The 6 is larger than 26, so it cannot be a letterThe first 0 cannot stand alone, and it cannot start a two-digit code, so no cut can beginThe string has an even length, so it cannot be cut into piecesThe base case dp[0] is 0, so every answer is 0

3. Predict the output of this program. It prints one line with three numbers. ``

cpp
#include <cstddef>
#include <cstdint>
#include <iostream>
#include <string_view>

std::uint64_t num_decodings_o1(std::string_view s) {
    std::uint64_t two_back = 1;  // dp[i - 2]
    std::uint64_t one_back = 1;  // dp[i - 1]
    for (std::size_t i = 1; i <= s.size(); ++i) {
        std::uint64_t current = 0;
        if (s[i - 1] != '0') current += one_back;
        if (i >= 2) {
            int two = (s[i - 2] - '0') * 10 + (s[i - 1] - '0');
            if (two >= 10 && two <= 26) current += two_back;
        }
        two_back = one_back;
        one_back = current;
    }
    return one_back;
}

int main() {
    std::cout << num_decodings_o1("226") << ' ' << num_decodings_o1("1020") << ' '
              << num_decodings_o1("1111") << '\n';
}

``

3 0 52 1 53 1 53 1 4

4. Which of these strings have 0 ways to decode? Pick all that apply.

Pick every answer that is true.

5. In the matrix method, which entry of M to the power n holds F(n)? Give the row and column as indexes, counting from 0.

Type your answer.

Fibonacci and decode ways