The segments we draw must be of equal length, corresponding to a specific step size \( k \) (number of steps between vertices).
For each step size \( k \), we need to determine if it is possible to form a perfect matching (non-overlapping segments covering all vertices). The number of such perfect matchings depends on the greatest common divisor (gcd) of \( k \) and 24.
When choosing a step size \( k \), the 24-gon is decomposed into \( \gcd(k, 24) \) cycles, each of length \( \frac{24}{\gcd(k, 24)} \). For a perfect matching to exist, each cycle must be of even length.
For each valid step size ( k ):
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If the cycle length is 2 (diameters), there is exactly 1 way to match the vertices.
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For other even cycle lengths, each cycle contributes a factor of 2 to the number of perfect matchings.
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( k = 1 ): ( \gcd(1, 24) = 1 ), cycle length 24, 2 matchings.
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( k = 2 ): ( \gcd(2, 24) = 2 ), cycle length 12, ( 2^2 = 4 ) matchings.
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( k = 3 ): ( \gcd(3, 24) = 3 ), cycle length 8, ( 2^3 = 8 ) matchings.
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( k = 4 ): ( \gcd(4, 24) = 4 ), cycle length 6, ( 2^4 = 16 ) matchings.
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( k = 5 ): ( \gcd(5, 24) = 1 ), cycle length 24, 2 matchings.
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( k = 6 ): ( \gcd(6, 24) = 6 ), cycle length 4, ( 2^6 = 64 ) matchings.
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( k = 7 ): ( \gcd(7, 24) = 1 ), cycle length 24, 2 matchings.
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( k = 8 ): ( \gcd(8, 24) = 8 ), cycle length 3 (invalid, no matchings).
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( k = 9 ): ( \gcd(9, 24) = 3 ), cycle length 8, ( 2^3 = 8 ) matchings.
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( k = 10 ): ( \gcd(10, 24) = 2 ), cycle length 12, ( 2^2 = 4 ) matchings.
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( k = 11 ): ( \gcd(11, 24) = 1 ), cycle length 24, 2 matchings.
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( k = 12 ): ( \gcd(12, 24) = 12 ), cycle length 2, 1 matching.
Summing these values: 2 + 4 + 8 + 16 + 2 + 64 + 2 + 0 + 8 + 4 + 2 + 1 = \(\boxed{113}\).
BPHO没报上，废了