Problem Analysis

Given (n) where (1 \le n \le 20), we need to count the number of ways to completely cover the interval ([1, n]) using segments where each segment connects two coprime numbers.

First, preprocess all coprime pairs ({a, b}) where (\gcd(a, b) = 1) and (a < b). Note that ({1, 1}) is excluded. When (n = 20), there are exactly 127 such pairs.

Key Observations

Consider the conditions under which Mirko can achieve a cmoplete coverage:

• If no selected pair contains the number 1, then all chosen pairs satisfy (a, b \ge 2). In this case, Slavko can simply choose (x = 2), and Mirko fails.
• If no selected pair contains the number (n), then all pairs satisfy (a, b < n). Slavko can choose (x = n), and Mirko fails again.
• If among the selected pairs, there exist two pairs ({a, b}) and ({c, d}) with (a < b < c < d), then Slavko can choose (x = c), and coverage remains incomplete.

For Mirko to succeed, none of these three conditions can occur.

Reduction to Interval Covering

Treat each coprime pair as a line segment from (a) to (b). The problem becomes:

Given a set of segments, count the number of ways to select a subset that completely covers ([1, n]).

Dynamic Programming Solution

Sort all segments by their right endpoint. Define (dp_{i,j}) as the number of ways to cover ([1, j]) using the first (i) segments. Initialize (dp_{0,1} = 1). The answer is (dp_{cnt, n}), where (cnt) is the total number of coprime pairs.

The transition for segment (i) with endpoints (L_i) and (R_i):

[dp_{i,j} = dp_{i,j} + dp_{i-1,j}] (skip segment (i))

[dp_{i,R_i} = dp_{i,R_i} + dp_{i-1,j}] where (L_i \le j) (extend coverage using segment (i))

[ \begin{cases} dp_{i,j}=\min(dp_{i,j}, dp_{i-1,j}) & \text{skip segment } i \ dp_{i,R_i}=\max(dp_{i,R_i}, dp_{i-1,j}) & L_i \le j \le R_i \end{cases} ]

The final answer is the sum of valid coverages modulo (10^9).

Implementation:

#include <bits/stdc++.h>
#define ll long long
#define reg register
#define F(i, a, b) for(reg int i = (a); i <= (b); ++i)
using namespace std;

inline int read();

const int MAXN = 150;
const int MOD = 1000000000;

int n, pairCount;
struct Segment {
    int left, right;
    bool operator<(const Segment& other) const {
        return right < other.right;
    }
} segments[MAXN];

namespace DP {
    ll dp[MAXN][25];
    
    void solve() {
        dp[0][1] = 1;
        F(i, 1, pairCount) {
            F(j, 1, n) {
                dp[i][j] = (dp[i][j] + dp[i-1][j]) % MOD;
                if (segments[i].left <= j) {
                    dp[i][segments[i].right] = (dp[i][segments[i].right] + dp[i-1][j]) % MOD;
                }
            }
        }
        printf("%lld\n", dp[pairCount][n]);
    }
}

int main() {
    n = read();
    F(i, 1, n) {
        F(j, i + 1, n) {
            if (__gcd(i, j) == 1) {
                ++pairCount;
                segments[pairCount] = {i, j};
            }
        }
    }
    sort(segments + 1, segments + pairCount + 1);
    DP::solve();
    return 0;
}

inline int read() {
    reg int x = 0;
    reg char c = getchar();
    while (!isdigit(c)) c = getchar();
    while (isdigit(c)) {
        x = (x << 3) + (x << 1) + (c ^ 48);
        c = getchar();
    }
    return x;
}

Memoized DFS Alternative

An alternative approach uses depth-first search with memoization. Define (dfs(last, pos, remaining)) as the number of ways to complete coverage where:

• (last) is the index of the last used segment
• (pos) is the current rightmost covered position
• (remaining) is how many more segments to select

#include <bits/stdc++.h>
#define ll long long
#define reg register
#define F(i, a, b) for(reg int i = (a); i <= (b); ++i)
using namespace std;

inline int read();

const int MAXN = 150;
const int MOD = 1000000000;

int n, pairCount;
struct Segment {
    int left, right;
} segments[MAXN];

ll memo[MAXN][25][MAXN];

ll search(int lastIdx, int currentPos, int remaining) {
    if (remaining == 0) return currentPos >= n;
    if (lastIdx >= pairCount) return 0;
    if (memo[lastIdx][currentPos][remaining] != -1) {
        return memo[lastIdx][currentPos][remaining];
    }
    
    ll result = 0;
    F(i, lastIdx + 1, pairCount) {
        if (segments[i].left <= currentPos) {
            int newPos = max(segments[i].right, currentPos);
            result = (result + search(i, newPos, remaining - 1)) % MOD;
        }
    }
    memo[lastIdx][currentPos][remaining] = result;
    return result;
}

int main() {
    n = read();
    F(i, 1, n) {
        F(j, i + 1, n) {
            if (__gcd(i, j) == 1) {
                ++pairCount;
                segments[pairCount] = {i, j};
            }
        }
    }
    
    memset(memo, -1, sizeof(memo));
    ll answer = 0;
    F(i, 1, pairCount) {
        answer = (answer + search(0, 1, i)) % MOD;
    }
    printf("%lld\n", answer);
    return 0;
}

inline int read() {
    reg int x = 0;
    reg char c = getchar();
    while (!isdigit(c)) c = getchar();
    while (isdigit(c)) {
        x = (x << 3) + (x << 1) + (c ^ 48);
        c = getchar();
    }
    return x;
}

Both solutions achieve the correct answer by systematically counting all valid segment selections that achieve complete coverage of the interval.