poj 1286 Necklace of Beads

Posted 2020-06-17

Necklace of Beads

题意:用三种颜色给长度为n(n < 24)的环状手镯涂色，若能通过旋转或翻转得到则表示为同一种，问不同种涂色方案为多少？

思路:纯粹的等价类计算问题；

重点:对旋转和翻转转化为置换操作；

旋转:对间隔的长度进行枚举，即0 <= i < n；这样循环节就为n/gcd(i,n);直接弄成3的幂次方求和即可；

翻转:分奇偶，再求出对称轴的个数和每种情况下循环节的个数即可；

上面求出的a+b只是不动点的个数总和，最后要除以总的置换的个数即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 namespace std;
#define rep0(i,l,r) for(int i = (l);i < (r);i++)
#define rep1(i,l,r) for(int i = (l);i <= (r);i++)
#define rep_0(i,r,l) for(int i = (r);i > (l);i--)
#define rep_1(i,r,l) for(int i = (r);i >= (l);i--)
#define MS0(a) memset(a,0,sizeof(a))
#define MS1(a) memset(a,-1,sizeof(a))
#define MSi(a) memset(a,0x3f,sizeof(a))
#define inf 0x3f3f3f3f
#define lson l, m, rt << 1
#define rson m+1, r, rt << 1|1
typedef __int64 ll;
template<typename T>
void read1(T &m)
{
    T x=0,f=1;char ch=getchar();
    while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
    while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}
    m = x*f;
}
template<typename T>
void read2(T &a,T &b){read1(a);read1(b);}
template<typename T>
void read3(T &a,T &b,T &c){read1(a);read1(b);read1(c);}
template<typename T>
void out(T a)
{
    if(a>9) out(a/10);
    putchar(a%10+‘0‘);
}
ll f[30],ans[30];
template<typename T>
int gcd(T a,T b){return b?gcd(b,a%b):a;}
int main()
{
    f[0] = 1;
    rep1(i,1,24) f[i] = f[i-1]*3;
    rep1(i,1,24){
        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;
    }
    int m;
    while(read1(m),m >= 0){
        out(ans[m]);puts("");
    }
    return 0;
}

View Code

 同类型的题:手镯操作一样，只是输入时先是颜色数后是珠子数；

#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 namespace std;
#define rep0(i,l,r) for(int i = (l);i < (r);i++)
#define rep1(i,l,r) for(int i = (l);i <= (r);i++)
#define rep_0(i,r,l) for(int i = (r);i > (l);i--)
#define rep_1(i,r,l) for(int i = (r);i >= (l);i--)
#define MS0(a) memset(a,0,sizeof(a))
#define MS1(a) memset(a,-1,sizeof(a))
#define MSi(a) memset(a,0x3f,sizeof(a))
#define inf 0x3f3f3f3f
#define lson l, m, rt << 1
#define rson m+1, r, rt << 1|1
typedef __int64 ll;
template<typename T>
void read1(T &m)
{
    T x=0,f=1;char ch=getchar();
    while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
    while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}
    m = x*f;
}
template<typename T>
void read2(T &a,T &b){read1(a);read1(b);}
template<typename T>
void read3(T &a,T &b,T &c){read1(a);read1(b);read1(c);}
template<typename T>
void out(T a)
{
    if(a>9) out(a/10);
    putchar(a%10+‘0‘);
}
ll f[40];
template<typename T>
int gcd(T a,T b){return b?gcd(b,a%b):a;}
int main()
{
    int s,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];
        else
            b = s/2*(f[s/2] + f[(s+2)/2]);
        out((a + b)/2/s);
        puts("");
    }
    return 0;
}

View Code