CF 1093A - Dice Rolling

Rating: 800
Tags: math
Solve time: 6m 28s
Verified: no

Solution

Problem Understanding

We are given a standard six-faced dice, but instead of the usual values 1 to 6, its faces contain the integers 2, 3, 4, 5, 6, and 7, all distinct. Each roll produces one of these numbers, and the score for a sequence of rolls is the sum of the visible faces.

For each query value $x$, we are asked to determine any number of rolls $k$ such that it is possible to obtain a total sum of exactly $x$ using exactly $k$ rolls. We are not required to construct the sequence of rolls, only to output a valid $k$.

The key interpretation is that once we fix $k$, the question becomes whether we can represent $x$ as a sum of $k$ integers, each chosen from ${2,3,4,5,6,7}$.

The constraints are small: $t \le 100$ and $x \le 100$. This immediately rules out any need for heavy combinatorics or search. Any solution that computes each answer in constant time per query is sufficient.

A subtle point is that many different values of $k$ can work for the same $x$. The problem does not ask for a minimum or maximum, only any valid one. That flexibility is what makes the construction simple.

There are no real edge cases involving impossibility, since the statement guarantees a solution exists for every query.

Approaches

A brute-force interpretation would be to try every possible number of rolls $k$, and for each $k$, check whether $x$ can be formed using $k$ numbers from 2 to 7. For a fixed $k$, this is equivalent to asking whether $x$ lies in the range $[2k, 7k]$, since 2 is the minimum contribution of a roll and 7 is the maximum.

If we brute-force over all $k$ up to, say, $x$, and for each $k$ perform a check, the complexity remains small given constraints, but it is unnecessary. The structure of the problem suggests a direct construction.

The key observation is that for any chosen $k$, the achievable sums form a continuous interval from $2k$ to $7k$. This interval has no gaps because we can adjust individual rolls in steps of 1 while staying within the allowed face values. Therefore, instead of searching over sequences, we only need to find any integer $k$ such that $2k \le x \le 7k$.

Rearranging, we want:

$$\frac{x}{7} \le k \le \frac{x}{2}$$

Since $k$ must be an integer, any integer in this range works. A simple constructive choice is to take the smallest valid $k$ that still allows reaching $x$, which is:

$$k = \left\lceil \frac{x}{7} \right\rceil$$

Once we fix this $k$, the remaining difference $x - 2k$ can always be distributed by increasing some rolls from 2 up to 7.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force over k and sequences	O(x · 6^k)	O(1)	Too slow
Interval construction	O(1) per query	O(1)	Accepted

Algorithm Walkthrough

Read the number of queries $t$. Each query gives a target sum $x$.
For each $x$, compute the smallest integer $k$ such that $7k \ge x$. This ensures that even if all rolls are 7, we can reach or exceed $x$.
Output this $k$ as the answer for the query.

The reason we pick the ceiling of $x/7$ is that it minimizes the number of rolls while guaranteeing enough total capacity to reach $x$. Any smaller $k$ would cap the maximum achievable sum below $x$, making it impossible.

Why it works

For any fixed number of rolls $k$, the set of achievable sums is exactly the interval $[2k, 7k]$. This is because each roll independently contributes any value from 2 to 7, and adjusting one roll by ±1 changes the total sum by exactly 1 while remaining valid. Therefore all integers in the range are reachable.

Choosing $k = \lceil x/7 \rceil$ guarantees $x \le 7k$. At the same time, since $k \ge 1$, we always have enough structure to represent $x$ by starting from all 2s and increasing some rolls. Thus $x \ge 2k$ also holds automatically for this construction range in the problem constraints, and any remaining gap can be filled by incrementing selected rolls up to 7.

Python Solution

import sys
input = sys.stdin.readline

def solve():
    t = int(input())
    for _ in range(t):
        x = int(input())
        k = (x + 6) // 7
        print(k)

if __name__ == "__main__":
    solve()

The code processes each query independently. The expression (x + 6) // 7 is the standard integer ceiling division for $x/7$, ensuring we always pick the smallest $k$ such that $7k \ge x$.

The implementation avoids any simulation of dice rolls. The logic directly computes the minimal feasible number of rolls, which is sufficient because the problem allows any valid configuration.

Worked Examples

Example 1

Input: $x = 13$

We compute $k = \lceil 13/7 \rceil = 2$

Step	x	k = ceil(x/7)	7k	Feasible?
1	13	2	14	Yes

With 2 rolls, we can achieve sums from 4 to 14, so 13 is reachable. For instance, 6 + 7 works.

This confirms that the construction is sufficient even when the target is close to the upper boundary.

Example 2

Input: $x = 37$

We compute $k = \lceil 37/7 \rceil = 6$

Step	x	k	7k	Feasible?
1	37	6	42	Yes

With 6 rolls, achievable sums range from 12 to 42. The target 37 lies inside this interval.

This demonstrates that even when $x$ is not a multiple of 7, the slack created by multiple rolls allows fine adjustment.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(t)	Each query is handled with a single arithmetic operation
Space	O(1)	Only a few integers are stored

The constraints allow up to 100 queries, so a linear scan is trivial. The solution runs in constant time per test and is far below the limits.

Test Cases

import sys, io

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

    import sys
    input = sys.stdin.readline

    t = int(sys.stdin.readline())
    for _ in range(t):
        x = int(sys.stdin.readline())
        k = (x + 6) // 7
        print(k)

    return output.getvalue().strip()

# provided samples
assert run("4\n2\n13\n37\n100\n") == "1\n2\n6\n15"

# custom cases
assert run("1\n7\n") == "1"
assert run("1\n8\n") == "2"
assert run("1\n14\n") == "2"
assert run("1\n1\n") == "1"
assert run("1\n100\n") == "15"

Test input	Expected output	What it validates
7	1	exact single-roll boundary
8	2	crossing first feasibility gap
14	2	upper boundary of small k
1	1	minimal constraint handling
100	15	large value scaling

Edge Cases

One important boundary is when $x$ is exactly divisible by 7. For example, $x = 14$. The algorithm gives $k = 2$. With two rolls, the maximum is $14$, achieved by $7 + 7$, confirming correctness at the upper boundary.

Another case is when $x$ is just above a multiple of 7, such as $x = 8$. Here $k = 2$. The range is $[4, 14]$, so 8 is achievable, for example $2 + 6$. This shows that even when one roll is insufficient, adding a second roll creates enough flexibility to fill the gap.

Finally, for very small values like $x = 2$, the formula gives $k = 1$. The only possible sum is exactly one roll, which must be 2, matching the requirement exactly.