CF 2072A - New World, New Me, New Array

Rating: 800
Tags: greedy, implementation, math
Solve time: 1m 11s
Verified: yes

Solution

Problem Understanding

We start with an array of n zeros. In one move, we can pick any position and overwrite it with any integer in the range [-p, p]. After performing some number of such overwrites, we want the total sum of the array to become exactly k, and we want to do this using as few moves as possible. Each move only affects one position, and it completely replaces the previous value at that position.

So the real decision is not about positions, but about how much total sum we can “inject” into the array using at most n assignments, each contributing a value between -p and p.

The constraints are small: n ≤ 50, |k| ≤ 2500, and p ≤ 50. This means even a direct case analysis or greedy formula is enough. Anything like DP over states is unnecessary, and even a brute-force over possible assignments per position would be bounded enough in theory, but overkill in practice.

A common failure case is forgetting that we can overwrite the same index multiple times. Since only the final value matters per index, what matters is how many indices we use, not how often we touch them.

Another subtle case is when k = 0. The correct answer is always zero operations because the array already sums to zero, regardless of p.

Approaches

The brute-force idea would try all ways of assigning values to up to n positions and check which combinations sum to k. This quickly explodes because each position has 2p + 1 choices, leading to (2p + 1)^n possibilities, which is far too large even for n = 50.

The key observation is that each operation contributes independently to the total sum. After x operations, the best we can do in magnitude is x * p, because each assignment can contribute at most p or -p. So the reachable sums after exactly x operations form the interval [-x p, x p]. Since we want exactly k, the only question is whether k lies in that range, and what the smallest such x is.

This reduces the problem to finding the minimum x such that x * p ≥ |k|, while also ensuring x ≤ n because we only have n positions to modify.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	exponential	O(1)	Too slow
Greedy bound check	O(1)	O(1)	Accepted

Algorithm Walkthrough

For each test case, read n, k, and p. We only care about absolute value of k because we can use negative assignments symmetrically.
If k is zero, return 0 immediately since no operations are needed and the initial array already satisfies the condition.
Compute need = |k|. This represents how much total magnitude we must build using operations.
Each operation contributes at most p to the total magnitude. So the minimum number of operations required is ceil(need / p).
If this number is greater than n, then even modifying every element once is not enough to reach the required sum, so the answer is -1.
Otherwise, output the computed value.

Why it works: each operation contributes independently to the total sum, and we are free to assign positive or negative values up to p. The only limiting factor is total capacity, which is linear in the number of operations. Since we can always choose values to exactly match the required sum as long as capacity allows, the ceiling bound is both necessary and sufficient.

Python Solution

import sys
input = sys.stdin.readline

def solve():
    t = int(input())
    for _ in range(t):
        n, k, p = map(int, input().split())

        if k == 0:
            print(0)
            continue

        need = abs(k)

        # minimum operations needed if each contributes at most p
        ops = (need + p - 1) // p

        if ops > n:
            print(-1)
        else:
            print(ops)

if __name__ == "__main__":
    solve()

The implementation directly follows the derived bound. The ceiling division (need + p - 1) // p avoids floating-point operations and ensures correctness for all integer cases. The check ops > n enforces the constraint that we cannot perform more than n independent assignments.

Worked Examples

Consider the case n = 5, k = 7, p = 2.

Step	need	p	ops formula	result
1	7	2	ceil(7/2)	4

Since 4 ≤ 5, the answer is 4.

Now consider n = 3, k = 10, p = 3.

Step	need	p	ops formula	result
1	10	3	ceil(10/3)	4

Here 4 > 3, so the answer is -1.

The first example shows that partial contributions per operation accumulate linearly. The second shows that even though values are individually large enough, we are constrained by the number of available positions.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(t)	Each test case is processed in constant time
Space	O(1)	No extra storage beyond variables

The solution comfortably handles up to t = 1000 cases within limits since each case is a few arithmetic operations.

Test Cases

import sys, io

def solve_io(data):
    sys.stdin = io.StringIO(data)
    input = sys.stdin.readline
    t = int(input())
    out = []
    for _ in range(t):
        n, k, p = map(int, input().split())
        if k == 0:
            out.append("0")
            continue
        need = abs(k)
        ops = (need + p - 1) // p
        out.append(str(ops if ops <= n else -1))
    return "\n".join(out)

# provided samples
assert solve_io("8\n21 100 10\n9 -420 42\n5 -7 2\n13 37 7\n10 0 49\n1 10 9\n7 -7 7\n20 31 1\n") == \
"10\n-1\n4\n6\n0\n-1\n1\n-1"

# custom cases
assert solve_io("1\n5 0 3\n") == "0"
assert solve_io("1\n5 1 10\n") == "1"
assert solve_io("1\n5 100 1\n") == "-1"
assert solve_io("2\n10 -25 5\n3 9 2\n") == "5\n-1"

Test input	Expected output	What it validates
k = 0	0	Already satisfied state
single op sufficient	1	minimal non-zero case
impossible case	-1	capacity bound violation
mixed cases	5, -1	correctness across scenarios

Edge Cases

When k = 0, the algorithm immediately returns zero without considering p, because no modification is needed. When p is large relative to k, the formula collapses to one operation, since a single assignment can absorb the entire sum. When p = 1, the number of operations becomes exactly |k|, and feasibility depends purely on whether |k| ≤ n. The ceiling division ensures no off-by-one errors in all these cases, since it correctly handles both divisible and non-divisible magnitudes.