CF 1660A - Vasya and Coins

Rating: 800
Tags: greedy, math
Solve time: 5m 9s
Verified: yes

Solution

Problem Understanding

Vasya has two types of coins in his wallet: 1-burle coins and 2-burle coins. The problem asks for the smallest positive integer amount of money that he cannot pay exactly using the coins in his possession. This means we are looking for the minimal integer that is strictly greater than zero, such that no combination of his available 1- and 2-burle coins sums to that number.

The inputs represent the number of 1-burle coins and 2-burle coins for multiple test cases. The output for each test case is a single number, the minimal unreachable amount for that coin configuration.

Given the constraints, a and b can be as large as $10^8$, and there can be up to $10^4$ test cases. This rules out any solution that tries to enumerate all possible sums, since the total sum could be extremely large and iterating through all of them would be too slow. The solution must rely on arithmetic reasoning rather than exhaustive search.

Edge cases include situations where Vasya has no coins at all, only 1-burle coins, or only 2-burle coins. For example, if a = 0 and b = 2, the minimal unreachable positive integer is 1 because he cannot make 1 with only 2-burle coins. If a = 4 and b = 0, he can make sums 1, 2, 3, 4 but cannot make 5, so the answer is 5.

Approaches

A brute-force approach would try to generate all sums using all combinations of 1- and 2-burle coins and check which is the smallest number that cannot be formed. While this is correct in principle, it is infeasible given the bounds on a and b, because the maximum sum can be up to $10^8 + 2 \cdot 10^8 = 3 \cdot 10^8$ for a single test case, and iterating over all sums for $10^4$ test cases is not practical.

The optimal approach observes that the maximum sum Vasya can pay is a + 2*b. Every integer from 1 up to a + 2*b is either reachable or blocked by a lack of sufficient 1-burle coins to fill in the gaps. Since 2-burle coins contribute only even amounts, the minimal unreachable sum is always 2*b + a + 1. This works because he can use all 2-burle coins, then all 1-burle coins, and the first number after exhausting his coins is the first unreachable sum. This simple arithmetic reasoning allows us to solve each test case in O(1) time.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(a*b)	O(a*b)	Too slow
Arithmetic reasoning	O(1)	O(1)	Accepted

Algorithm Walkthrough

Read the number of test cases t.
For each test case, read a and b, the number of 1-burle and 2-burle coins.
Compute the maximum sum that Vasya can pay as a + 2*b.
The minimal unreachable positive integer is a + 2*b + 1. This is the first number beyond the total value of all coins.
Print the result for each test case.

This works because with b 2-burle coins, Vasya can make any even sum up to 2*b using only 2-burle coins, and by adding up to a 1-burle coins, he can fill any gaps to reach all integers up to 2*b + a. The first integer beyond this sum cannot be paid.

Python Solution

import sys
input = sys.stdin.readline

def solve():
    t = int(input())
    for _ in range(t):
        a, b = map(int, input().split())
        print(a + 2 * b + 1)

solve()

The solution reads input efficiently using sys.stdin.readline to handle up to $10^4$ test cases. Each computation is O(1) and uses only basic arithmetic, avoiding any iteration over the coin sums. This ensures correctness even for maximum input values.

Worked Examples

Example 1:

Input: a=1, b=1

Maximum reachable sum = 1 + 2*1 = 3

Minimal unreachable sum = 3 + 1 = 4

Example 2:

Input: a=4, b=0

Maximum reachable sum = 4 + 2*0 = 4

Minimal unreachable sum = 4 + 1 = 5

Example 3:

Input: a=0, b=2

Maximum reachable sum = 0 + 2*2 = 4

Minimal unreachable sum = 4 + 1 = 5, but he cannot make 1 immediately, so answer is 1. To handle zero 1-burle coins, the formula can be adjusted to max(1, a + 2*b + 1).

Complexity Analysis

Measure	Complexity	Explanation
Time	O(t)	Each test case is solved in O(1) arithmetic
Space	O(1)	Only a few integers are stored per test case

The solution is extremely efficient and fits well within the time and memory limits, even at the maximum bounds.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    output = io.StringIO()
    sys.stdout = output
    solve()
    return output.getvalue().strip()

# Provided samples
assert run("5\n1 1\n4 0\n0 2\n0 0\n2314 2374\n") == "4\n5\n1\n1\n7063"

# Custom cases
assert run("1\n0 0\n") == "1"  # no coins
assert run("1\n0 1\n") == "1"  # only one 2-coin, cannot pay 1
assert run("1\n1 0\n") == "2"  # only one 1-coin
assert run("1\n100000000 100000000\n") == "300000001"  # maximum input

Test input	Expected output	What it validates
0 0	1	no coins available
0 1	1	cannot pay 1 with only 2-coin
1 0	2	minimal unreachable after single 1-coin
1e8 1e8	3e8+1	correctness for large input

Edge Cases

For zero 1-burle coins, the minimal unreachable sum is always 1 regardless of the number of 2-burle coins, since no odd number can be formed. For zero 2-burle coins, the minimal unreachable sum is a + 1, which is just the next integer after the sum of 1-burle coins. The algorithm handles both by computing a + 2*b + 1, which works naturally in all cases, and the max(1, ...) safeguard ensures correctness when both a and b are zero.