CF 54C - First Digit Law

Rating: 2000
Tags: dp, math, probabilities
Solve time: 1m 10s
Verified: yes

Solution

Problem Understanding

We are given a set of N random integers, where each integer i can take any value in a given inclusive range [L_i, R_i], all values equally likely. The task is to compute the probability that at least K% of these N integers start with the digit 1.

In other words, for each possible set of values chosen from their ranges, we examine the most significant digit (MSD) of each value and count how many are 1. If that count is at least ceil(K * N / 100), the set is considered "good." We want the probability that a randomly chosen set of values is good.

The constraints reveal important characteristics of the problem. N can go up to 1000, and the ranges [L_i, R_i] can extend up to 10^18. This immediately rules out any approach that enumerates all numbers in the ranges because the total number of combinations can be astronomically large. Even iterating over each integer in a range is impossible for large segments, so we must reason in terms of first-digit probabilities rather than raw enumeration.

There are subtle edge cases here. For example, if a range [1, 9] contains only numbers with first digit 1 for a fraction of the numbers, we must calculate this fraction precisely. Small ranges that cross powers of ten, like [95, 105], are tricky because the numbers starting with 1 are only [100, 105]. A naive approach might assume uniform distribution of first digits or ignore boundaries, which produces incorrect results.

Approaches

The brute-force approach would enumerate every combination of N numbers and count how many have a first digit of 1. For each variable i, there are R_i - L_i + 1 possibilities. The total number of sets is the product of all these lengths. Even for N = 10 with ranges of size 10^6, the product exceeds 10^60, making brute-force impossible.

The key insight is that we do not need the actual numbers, only the probability that a chosen number starts with 1. For each range [L_i, R_i], we can compute p_i, the probability that a number drawn uniformly from this range starts with 1. Computing p_i efficiently is non-trivial but feasible by counting intervals of numbers with MSD 1. Once we have these p_i values, the problem reduces to computing the probability that at least X of N independent events with probabilities p_i occur, where X = ceil(K * N / 100). This is a classic dynamic programming problem, similar to subset sum but with probabilities.

We define dp[j] as the probability that exactly j of the first i numbers have first digit 1. For each variable i and each possible count j, we update dp[j] using p_i and 1-p_i. This approach is linear in N times N for the DP, i.e., O(N^2), which is feasible for N <= 1000.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(product of ranges)	O(N)	Impossible
DP with first-digit probabilities	O(N^2)	O(N)	Accepted

Algorithm Walkthrough

For each variable i, compute p_i, the probability that a number chosen uniformly from [L_i, R_i] has first digit 1. To do this efficiently, iterate over powers of ten. For a given k, the numbers starting with 1 are [10^k, 2*10^k - 1]. Intersect this interval with [L_i, R_i] and sum lengths, then divide by the total range length R_i - L_i + 1.
Initialize a DP array dp of size N+1 with dp[0] = 1.0 and all others 0.0. Here, dp[j] represents the probability that exactly j of the first i numbers have first digit 1.
Iterate over each i from 1 to N. For each j from i down to 0, update dp[j] using dp[j-1]*p_i (if j>0) plus dp[j]*(1-p_i) to account for including or not including the current number in the count.
After processing all N variables, sum dp[j] for all j >= X, where X = ceil(K * N / 100). This sum is the required probability.
Print the probability with high precision.

Why it works: the DP invariant is that after processing i variables, dp[j] exactly represents the probability that exactly j out of the first i numbers have first digit 1. At each step, probabilities are combined using the law of total probability for independent events.

Python Solution

import sys
import math
input = sys.stdin.readline

def first_digit_probability(L, R):
    count = 0
    pow10 = 1
    while pow10 <= R:
        start = max(L, pow10)
        end = min(R, pow10 * 2 - 1)
        if start <= end:
            count += end - start + 1
        pow10 *= 10
    return count / (R - L + 1)

def solve():
    N = int(input())
    intervals = []
    for _ in range(N):
        L, R = map(int, input().split())
        intervals.append((L, R))
    K = int(input())
    X = math.ceil(K * N / 100)
    
    p = [first_digit_probability(L, R) for L, R in intervals]
    
    dp = [0.0] * (N + 1)
    dp[0] = 1.0
    
    for i in range(N):
        for j in range(i, -1, -1):
            dp[j+1] = dp[j+1] + dp[j] * p[i] if j >= 0 else dp[j+1]
            dp[j] = dp[j] * (1 - p[i])
    
    result = sum(dp[X:])
    print(f"{result:.15f}")

solve()

The function first_digit_probability carefully counts numbers starting with 1 within [L, R] by intersecting with intervals [10^k, 2*10^k-1]. The DP update uses a reverse loop to avoid overwriting values that are still needed. This is a classic trick when building subset-count probabilities in-place.

Worked Examples

Sample Input 1

1
1 2
50

Step	dp array	Explanation
Initial	[1.0, 0.0]	No numbers processed
Process 1st interval p=0.5	[0.5, 0.5]	Probability 0.5 of first digit 1
Sum dp[j >= 1]	0.5	Threshold X=ceil(50*1/100)=1

This shows that the algorithm correctly identifies the probability that at least 50% (here 1 out of 1) has first digit 1.

Custom Input

Step	dp array	Explanation
Initial	[1.0, 0.0, 0.0]
Process first interval p=0.1	[0.9,0.1,0.0]	1 number processed
Process second interval p=0.1	[0.81,0.18,0.01]	2 numbers processed
Threshold X=ceil(50*2/100)=1	sum(dp[1:])=0.19	Correct probability

Complexity Analysis

Measure	Complexity	Explanation
Time	O(N * logR_max + N^2)	Computing first-digit probabilities takes O(logR) per variable; DP takes O(N^2)
Space	O(N)	DP array stores N+1 probabilities

For N=1000, O(N^2)=10^6 operations, and logR_max ≤ 60, so this fits comfortably under the 2s limit.

Test Cases

import sys, io

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

# Provided sample
assert run("1\n1 2\n50\n") == "0.500000000000000"

# Minimum input
assert run("1\n1 1\n0\n") == "1.000000000000000"

# Maximum K
assert run("2\n1 10\n10 20\n100\n") == "0.010000000000000"

# All equal numbers
assert run("3\n1 1\n1 1\n1