CF 104833F - 舌切雀（Easy Version）

CF 104833F - \u820c\u5207\u96c0\uff08Easy Version\uff09

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Solve time: 42s
Verified: yes

Solution

Problem Understanding

We are given a function defined on natural numbers. For a number $x$, we look at whether there exists an integer exponent $k > 1$ such that $x^k$ is a rational number. The function $f(x)$ outputs 1 if such an exponent exists, otherwise it outputs 0. For each test case, we are asked to compute the sum of $f(i)$ over all integers $i$ from 1 to $n$.

The input contains multiple queries, and each query provides a single value $n$ up to $10^{12}$. This immediately suggests that any solution that iterates over all values up to $n$ is impossible, since even a single test case could require up to $10^{12}$ operations.

The key subtlety in this statement is that we are evaluating integer inputs, and the condition involves powers of those integers. A common pitfall is to overthink the condition as something involving deep number theory or non-integer inputs. However, since every $i$ is a natural number, every power $i^k$ is also an integer, and thus automatically a rational number.

A naive misinterpretation would be to attempt checking whether there exists some exponent making the result rational via complicated algebraic reasoning. That is unnecessary here and would lead to overcomplicated implementations that still miss the core simplification.

A typical edge case to test intuition is $i = 1$. Since $1^k = 1$, it is always rational, so $f(1) = 1$. For any other integer, say $i = 2$, we still have $2^k$ being an integer for all $k$, hence rational, so $f(2) = 1$. This pattern holds for every natural number.

So every term in the sum contributes 1.

Approaches

A brute-force approach would directly evaluate the definition for each $i$ from 1 to $n$, and for each such $i$, try different values of $k > 1$ and check whether $i^k$ is rational. Since $i^k$ can be computed exactly for integers, this approach would always confirm the condition. However, even if we restrict ourselves to a small range of $k$, we are still performing $O(n)$ work per query, which is far beyond feasible when $n$ can reach $10^{12}$.

The key observation is that the condition does not actually filter out any natural number. For any integer $i \ge 1$, and any integer exponent $k > 1$, the value $i^k$ remains an integer and therefore belongs to the set of rational numbers. This means the existential condition in the definition of $f(x)$ is always satisfied.

Once we realize that every element contributes 1, the problem reduces to computing a simple prefix sum.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	$O(nT)$	$O(1)$	Too slow
Optimal	$O(T)$	$O(1)$	Accepted

Algorithm Walkthrough

1. Read the number of test cases $T$. Each test case is independent and asks for a range sum up to $n$.
2. For each test case, read the integer $n$.
3. Recognize that every integer $i$ in $[1, n]$ satisfies $f(i) = 1$, so the required sum is simply the count of integers in the range.
4. Compute the answer as $n$.
5. Output the result.

Why it works

The function $f(i)$ depends on whether there exists an exponent $k > 1$ such that $i^k$ is rational. Since $i$ is a natural number, any exponentiation $i^k$ remains an integer. Every integer is a rational number, so the condition is always true. This makes $f(i)$ identically equal to 1 over the entire domain $[1, n]$, ensuring the sum is exactly $n$.

Python Solution

import sys
input = sys.stdin.readline

def solve():
    t = int(input())
    for _ in range(t):
        n = int(input())
        print(n)

if __name__ == "__main__":
    solve()

The solution simply reads each query and outputs $n$, since every term contributes exactly one unit to the sum.

The only subtle implementation detail is ensuring fast I/O because $T$ can be as large as $2 \times 10^3$, but even standard input/output is sufficient here due to constant-time processing per test case.

Worked Examples

Consider an input with multiple values of $n$.

For $n = 1$, we sum over a single value. Since $f(1) = 1$, the result is 1.

For $n = 5$, each of the values 1 through 5 contributes 1.

i	f(i)
1	1
2	1
3	1
4	1
5	1

The sum is 5.

This confirms that the function does not introduce any variation across the range.

Complexity Analysis

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

The constraints allow up to 2000 test cases and values up to $10^{12}$, but since each query is answered immediately, the solution easily fits within limits.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    import sys
    input = sys.stdin.readline

    def solve():
        t = int(input())
        for _ in range(t):
            n = int(input())
            print(n)

    old_stdout = sys.stdout
    sys.stdout = io.StringIO()
    solve()
    out = sys.stdout.getvalue().strip()
    sys.stdout = old_stdout
    return out

# provided sample-style checks
assert run("1\n1\n") == "1"
assert run("2\n1\n5\n") == "1\n5"

# custom cases
assert run("3\n10\n100\n1000\n") == "10\n100\n1000"
assert run("1\n1000000000000\n") == "1000000000000"
assert run("4\n2\n3\n4\n5\n") == "2\n3\n4\n5"

Test input	Expected output	What it validates
single small n	1	base case correctness
multiple queries	n values	independence of queries
large n	n	handling large constraints
consecutive range	same numbers	no hidden transformation

Edge Cases

For $n = 1$, the loop contains only a single element, and since $f(1)$ is always 1, the output is 1. The algorithm handles this directly without any special branching.

For very large $n$, such as $10^{12}$, the computation does not depend on iterating over the range, so the same constant-time logic applies. The output remains $n$, confirming that no overflow or approximation is involved since Python integers handle arbitrary size safely.