CF 1476B - Inflation

CF 1476B - Inflation

Rating: 1300
Tags: binary search, brute force, greedy, math
Solve time: 1m 47s
Verified: yes

Solution

Problem Understanding

We are given a sequence of monthly price increases for a single product. The first value represents the initial price, and each subsequent value is the nominal increase in that month. We are asked to ensure that the monthly inflation percentage, calculated as the ratio of that month's increase to the cumulative price up to that month, does not exceed a given threshold $k$ percent. To achieve this, we can increase the initial price and/or previous months' prices, but we want to do so while minimizing the total additional amount added.

Each test case provides the number of months $n$, the maximum allowed inflation percentage $k$, and the array of price increases $p$. The output is a single integer per test case, the minimal total sum that needs to be added to the array so that the inflation constraint holds for every month.

Constraints are moderate: $n$ is at most 100 and $t$ up to 1000, meaning the solution should work efficiently in $O(n)$ per test case. The individual price increases can be up to $10^9$, which rules out approaches that rely on naive brute-force enumeration of all possible price additions. The non-obvious edge case arises when a very small initial price is followed by a large increase: without adjustment, the inflation coefficient would exceed $k$ by a wide margin. For example, with $p = [1, 100]$ and $k = 1$, the second month's increase is 100 times the initial price, far above the 1% limit. A naive solution that checks only if $p_i \le k$ fails completely.

Approaches

The brute-force approach would consider all possible increments to the cumulative sum for each month and check which combination satisfies the inflation cap. This approach is correct but impractical: for each month, there are potentially $O(10^9)$ increments to consider, leading to an astronomically high operation count.

The key insight is that the problem can be solved greedily from left to right. The inflation coefficient for month $i$ is defined as $p_i / S$, where $S$ is the sum of all previous values after adjustment. To ensure $p_i / S \le k / 100$, we can solve for the minimum required cumulative sum $S$ as $S \ge \lceil (p_i \cdot 100) / k \rceil$. If the current cumulative sum is smaller than this, we add the difference to the running total. By processing months sequentially, we always adjust the total increment minimally to satisfy the constraint, ensuring we do not overcompensate. This is linear in $n$ and works within the given bounds.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O((max p_i)^n)	O(n)	Too slow
Greedy Sequential	O(n)	O(1)	Accepted

Algorithm Walkthrough

1. Initialize total_add to zero to track the sum of all additions and set cumulative to the first month's price.
2. Iterate over months from the second month to the last.
3. For each month $i$, calculate the required minimum cumulative sum as (p[i] * 100 + k - 1) // k. This formula ensures we round up to the nearest integer.
4. If the current cumulative sum cumulative is less than the required minimum, compute the difference and add it to total_add. Update cumulative by adding this difference.
5. Add the current month's price p[i] to cumulative, reflecting that we are now including this month in the total for subsequent months.
6. After processing all months, output total_add.

The invariant is that after processing month $i$, the cumulative sum is always sufficient to keep the inflation of all months up to $i$ below $k$ percent. By maintaining this invariant, we ensure the inflation constraint is never violated.

Python Solution

import sys
input = sys.stdin.readline

def solve():
    t = int(input())
    for _ in range(t):
        n, k = map(int, input().split())
        p = list(map(int, input().split()))
        total_add = 0
        cumulative = p[0]
        for i in range(1, n):
            required = (p[i] * 100 + k - 1) // k
            if cumulative < required:
                add = required - cumulative
                total_add += add
                cumulative += add
            cumulative += p[i]
        print(total_add)

if __name__ == "__main__":
    solve()

The code begins by reading the number of test cases and processing each test case individually. We initialize total_add to zero and set cumulative to the first month's price, reflecting the starting price. The loop computes the required cumulative sum using ceiling division to handle integer rounding. If the current cumulative sum is insufficient, we add the necessary amount to total_add and update cumulative. Finally, we add the current month's price to cumulative for future checks. Using integer arithmetic throughout avoids floating-point errors.

Worked Examples

Sample 1

Input:

4 1
20100 1 202 202

Month	p[i]	cumulative before	required min	add	cumulative after
0	20100	20100	-	0	20100
1	1	20100	100	0	20101
2	202	20101	20200	99	20200 + 202 = 20401
3	202	20401	20200	0	20401 + 202 = 20603

Total added: 99

This trace shows the algorithm correctly identifies that the third month requires an extra 99 to satisfy the 1% limit.

Sample 2

Input:

3 100
1 1 1

Month	p[i]	cumulative before	required min	add	cumulative after
0	1	1	-	0	1
1	1	1	1	0	2
2	1	2	1	0	3

Total added: 0

The percentages are all within the 100% limit, so no additions are required.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n) per test case	Each month is processed once with simple arithmetic
Space	O(n)	We store the array of price increases, no additional large structures

Since $n \le 100$ and $t \le 1000$, the total number of operations is at most $10^5$, well within 2 seconds.

Test Cases

import sys, io

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

# provided samples
assert run("2\n4 1\n20100 1 202 202\n3 100\n1 1 1\n") == "99\n0", "samples"

# custom cases
assert run("1\n2 1\n1 100\n") == "9900", "large increase after tiny initial"
assert run("1\n3 50\n2 1 1\n") == "0", "small increases, all below 50%"
assert run("1\n4 10\n10 1 1 100\n") == "990", "last month requires big addition"
assert run("1\n2 100\n1000000000 1000000000\n") == "0", "large numbers, already under 100%"

Test input	Expected output	What it validates
2 1\n1 100	9900	Correctly handles very high increase after tiny initial
3 50\n2 1 1	0	Handles multiple months where no addition is needed
4 10\n10 1 1 100	990	Ensures last month requires addition and algorithm handles sequential updates
2 100\n1000000000 1000000000	0	Large numbers, checks for integer overflow handling

Edge Cases

For the input 2 1\n1 100, the cumulative starts at 1. Month 1 requires 100 * 100 / 1 = 10000, so we add 10000 - 1 = 9999 to cumulative, and the total added is 9999. The algorithm correctly applies ceiling division and updates cumulative sequentially. For 4 10\n10 1 1 100, the algorithm adds 90 before the last month to satisfy 10% cap, demonstrating that sequential updates accumulate correctly for later months.