CF 104804A - Ресурсы

CF 104804A - \u0420\u0435\u0441\u0443\u0440\u0441\u044b

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Solve time: 1m 2s
Verified: yes

Solution

Problem Understanding

Each test case describes a collection of resource types gathered during a board game session. For every type, we are given how many units of that resource were collected and how many points a single unit is worth. The task is to compute the total score by aggregating contributions from all resource types.

In more concrete terms, each line gives a pair of numbers. The first number is a count of items of a specific kind, and the second number is the value of one such item. The contribution of that type to the final score is simply their product, and the answer is the sum of these contributions over all types.

The constraints allow up to 100000 resource types, and each value can be as large as 100000. This immediately rules out any solution that attempts anything beyond a single linear pass over the input. Even an O(n log n) approach is unnecessary because there is no structure to exploit beyond direct aggregation. The only viable approach is to accumulate the sum as we read the input.

A subtle issue that can arise in careless implementations is integer overflow in languages with fixed-width integers. For example, if all inputs are at their maximum, a single product can reach 10^10, and summing up to 10^5 such values can reach 10^15. In Python this is safe, but in other languages this requires 64-bit integers.

Another potential mistake is trying to separate accumulation into two arrays and processing them later unnecessarily, which adds overhead but does not change correctness. Since each line is independent, delaying computation provides no benefit.

Approaches

A brute-force interpretation would be to expand every resource unit individually and then assign values one by one. That means for each type, we conceptually create a list of size a_i and assign b_i to each element. This is correct but immediately becomes infeasible when a_i is large, because the total number of expanded elements could reach 10^10 in the worst case.

The key observation is that all units of the same type are identical in value contribution. There is no interaction between resource types, and no ordering or constraints across them. This collapses the problem into computing a weighted sum where each weight is simply a_i and each value is b_i. Instead of expanding, we multiply and accumulate directly.

This reduces the entire task to a single pass over the input.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force Expansion	O(∑a_i)	O(∑a_i)	Too slow
Direct Summation	O(n)	O(1)	Accepted

Algorithm Walkthrough

1. Initialize a variable total to zero. This will store the running sum of all contributions.
2. Read the number of resource types n.
3. For each of the next n lines, read a pair (a_i, b_i).
4. Compute the contribution of this type as a_i * b_i. This represents the total value of all items of that type combined.
5. Add this contribution to total.
6. After processing all lines, output total.

The key design choice is performing multiplication at read time instead of storing values. This avoids unnecessary memory usage and ensures the computation is strictly linear in the number of input lines.

Why it works

Each resource type is independent of all others, and every unit within a type has identical value. Therefore, the total score is the sum of disjoint groups, where each group contributes exactly a_i * b_i. Since addition is associative and commutative, accumulating contributions incrementally preserves correctness regardless of input order. No interaction terms exist, so no additional state is required beyond the running sum.

Python Solution

import sys
input = sys.stdin.readline

def solve():
    n = int(input())
    total = 0
    for _ in range(n):
        a, b = map(int, input().split())
        total += a * b
    print(total)

if __name__ == "__main__":
    solve()

The solution reads each pair once and immediately folds it into the answer. The multiplication is done before addition to ensure we never need to store intermediate expanded representations. Using Python’s arbitrary precision integers guarantees correctness even when the sum exceeds 64-bit limits.

Worked Examples

Sample 1

Input:

5
1 3
2 6
3 1
4 5
5 10

Step	a_i	b_i	Contribution	Total
1	1	3	3	3
2	2	6	12	15
3	3	1	3	18
4	4	5	20	38
5	5	10	50	88

The final sum is 88, which matches the expected output. This trace shows how each type independently contributes and is accumulated without interference.

Sample 2

Input:

5
0 1
0 3
4 4
2 7
1 1

Step	a_i	b_i	Contribution	Total
1	0	1	0	0
2	0	3	0	0
3	4	4	16	16
4	2	7	14	30
5	1	1	1	31

This example highlights that zero counts effectively skip contributions, and the algorithm naturally handles them without special cases.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n)	Each resource type is processed exactly once with constant-time arithmetic
Space	O(1)	Only a single accumulator variable is used

The constraints allow up to 100000 entries, and a linear scan with simple arithmetic is well within typical limits. No additional structures or preprocessing are required.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    from __main__ import solve
    from contextlib import redirect_stdout
    out = io.StringIO()
    with redirect_stdout(out):
        solve()
    return out.getvalue().strip()

# provided samples
assert run("""5
1 3
2 6
3 1
4 5
5 10
""") == "88"

assert run("""5
0 1
0 3
4 4
2 7
1 1
""") == "31"

# custom cases
assert run("""1
0 100
""") == "0"

assert run("""3
100000 100000
0 5
1 1
""") == "10000000001"

assert run("""4
1 1
1 1
1 1
1 1
""") == "4"

assert run("""2
99999 99999
1 0
""") == "9999800001"

Test input	Expected output	What it validates
single zero pair	0	zero contribution handling
large values	10000000001	overflow scale correctness
uniform small values	4	basic accumulation correctness
mixed zero multiplier	9999800001	ignoring zero-value terms

Edge Cases

One edge case is when all resource counts are zero. For input:

3
0 10
0 20
0 30

each contribution is zero, so the running sum stays at zero throughout execution. The algorithm performs three multiplications but all results are zero, and the final output remains 0 without any special branching.

Another case is when only one type exists:

1
12345 67890

The algorithm performs a single multiplication and outputs the result directly. No initialization issues arise because the accumulator starts at zero, and the single update correctly captures the full answer.

A third case is when values are at maximum magnitude. Even though intermediate products can be large, Python handles arbitrary precision naturally, so repeated accumulation does not overflow or lose precision.