UVa 10294 (Pólya count) Arif in Dhaka (First love part 2)

Burnside theorem: If a coloring scheme s is changed after substitution F, it is called S is the fixed point of F, the number of fixed points of displacement F is recorded as C (f). The number of equivalence classes equals the average of all C (f).

A necklace, a bracelet, the difference is that one can flip a can't, use T color dye n beads, the number of equivalence classes.

Rotation permutation group There are total n permutations, respectively, corresponding to the whole necklace counterclockwise rotation 0, one, 2 ... The substitution of beads.

For the first permutation, No. 0, I, 2i ... A bead consists of a cycle of GCD (n, i) loops with N/GCD (n, i) beads in each cycle.

So n permutations, the fixed points of each permutation have tgcd (i, N).

For a reversed permutation group, there are two conditions based on the odd-even division of N:

• n is odd, the symmetric axis has n, each passes through a bead, each permutation has (n+1)/2 loops, including (n-1)/2 2-length loops and a 1-length cycle (that is, the bead that crosses). So the total number of fixed points is NT (n+1)/2
• n is even, there are n/2 through the beads of the symmetrical axis, each axis of symmetry through two beads. There are a total of n/2+1 loops, including n/2-1 with a 2-length loop and 2 1-length loops, and a symmetric axis with N/2 that does not pass through the beads, with a total of N/2 2-length loops.

Another A = sum{tgcd (M, i) | 0≤i≤n-1}

If n is odd, B = NT (n+1)/2

If n is even, B = n/2 * (t (n/2+1) + TN/2)

The answers are a/n and (a+b)/(2n) respectively.

1#include <cstdio>2typedefLong LongLL;3 4 intgcdintAintb)5{returnb = =0? A:GCD (b, a%b); }6 7 Const intMAXN = -;8 LL P[MAXN];9 Ten intMain () One { A     //freopen ("In.txt", "R", stdin); -  -p[0] =1; the     intN, t; -      while(SCANF ("%d%d", &n, &t) = =2) -     { -          for(inti =1; I <= N; i++) P[i] = p[i-1] *T; +LL A =0, B =0; -          for(inti =0; I < n; i++) A + =P[GCD (n, i)]; +         if(N &1) b = n * p[(n+1)/2]; A         Elseb = n/2* (p[n/2+1] + p[n/2]); atprintf"%lld%lld\n", a/n, (a+b)/n/2); -     } -  -     return 0; -}

code June

UVa 10294 (Pólya count) Arif in Dhaka (First love part 2)