POJ 1286 Necklace of beads

Necklace of Beads

Test instructions: In three colors to the length of N (n < 24) ring bracelet coloring, if can be rotated or flipped to be expressed as the same type, ask how many different coloring scheme?

Idea: Pure equivalence class calculation problem;

Emphasis: Conversion of rotation and rollover to displacement operations;

Rotation: The length of the interval is enumerated, that is 0 <= i < n, so that the cyclic section is N/GCD (i,n), directly into the power of 3 to sum;

Flip: The odd even, and then find the number of symmetric axis and the number of cycles in each case can be;

The above-calculated a+b is just the sum of the number of fixed points, the final divided by the total number of permutations that is 2n;

#include <iostream>#include<cstdio>#include<cstring>#include<string.h>#include<algorithm>#include<vector>#include<cmath>#include<stdlib.h>#include<time.h>#include<stack>#include<Set>#include<map>#include<queue>using namespacestd;#defineRep0 (I,L,R) for (int i = (l); i < (R); i++)#defineREP1 (I,L,R) for (int i = (l); I <= (r); i++)#defineRep_0 (i,r,l) for (int i = (r); i > (l); i--)#defineRep_1 (i,r,l) for (int i = (r); I >= (l); i--)#defineMS0 (a) memset (A,0,sizeof (a))#defineMS1 (a) memset (A,-1,sizeof (a))#defineMSi (a) memset (A,0x3f,sizeof (a))#defineINF 0x3f3f3f3f#defineLson L, M, RT << 1#defineRson m+1, R, RT << 1|1typedef __int64 Ll;template<typename t>voidRead1 (T &m) {T x=0, f=1;CharCh=GetChar ();  while(ch<'0'|| Ch>'9'){if(ch=='-') f=-1; ch=GetChar ();}  while(ch>='0'&&ch<='9') {x=x*Ten+ch-'0'; ch=GetChar ();} M= x*F;} Template<typename t>voidRead2 (T &a,t &b) {Read1 (a); Read1 (b);} Template<typename t>voidRead3 (T &a,t &b,t &B) {Read1 (a); Read1 (b); Read1 (c);} Template<typename t>void  out(T a) {if(a>9) out(ATen); Putchar (A%Ten+'0');} ll f[ -],ans[ -];template<typename t>intGCD (T a,t b) {returnB?GCD (b,a%b): A;}intMain () {f[0] =1; REP1 (i,1, -) F[i] = f[i-1]*3; REP1 (i,1, -) {ll a=0, B =0; Rep0 (J,0, i) A + =F[GCD (i,j)]; if(I &1) b = i*f[(i+1)/2]; Else{b= i/2* (f[i/2] + f[(i+2)/2]); } Ans[i]= (A + b)/2/i; }    intm;  while(Read1 (m), M >=0){         out(Ans[m]);p UTS (""); }    return 0;}

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The same type of title: Bracelet operation, just enter the number of colors after the first number of beads;

#include <iostream>#include<cstdio>#include<cstring>#include<string.h>#include<algorithm>#include<vector>#include<cmath>#include<stdlib.h>#include<time.h>#include<stack>#include<Set>#include<map>#include<queue>using namespacestd;#defineRep0 (I,L,R) for (int i = (l); i < (R); i++)#defineREP1 (I,L,R) for (int i = (l); I <= (r); i++)#defineRep_0 (i,r,l) for (int i = (r); i > (l); i--)#defineRep_1 (i,r,l) for (int i = (r); I >= (l); i--)#defineMS0 (a) memset (A,0,sizeof (a))#defineMS1 (a) memset (A,-1,sizeof (a))#defineMSi (a) memset (A,0x3f,sizeof (a))#defineINF 0x3f3f3f3f#defineLson L, M, RT << 1#defineRson m+1, R, RT << 1|1typedef __int64 Ll;template<typename t>voidRead1 (T &m) {T x=0, f=1;CharCh=GetChar ();  while(ch<'0'|| Ch>'9'){if(ch=='-') f=-1; ch=GetChar ();}  while(ch>='0'&&ch<='9') {x=x*Ten+ch-'0'; ch=GetChar ();} M= x*F;} Template<typename t>voidRead2 (T &a,t &b) {Read1 (a); Read1 (b);} Template<typename t>voidRead3 (T &a,t &b,t &c) {Read1 (a); Read1 (b); Read1 (c);} Template<typename t>void  out(T a) {if(a>9) out(ATen); Putchar (A%Ten+'0');} ll f[ +];template<typename t>intGCD (T a,t b) {returnB?GCD (b,a%b): A;}intMain () {intS,c;  while(Read2 (c,s), S +c) {f[0] =1; REP1 (i,1, s) f[i] = f[i-1]*C; ll a=0, B =0; Rep0 (J,0, s) A + =F[GCD (s,j)]; if(S &1) b = s*f[(s+1)/2]; Elseb= s/2* (f[s/2] + f[(s+2)/2]);  out((A + B)/2/s); Puts (""); }    return 0;}

View Code

POJ 1286 Necklace of beads