POJ 2409 Let it Bead

迈不过友情╰ 2022-05-15 07:15 112阅读 0赞

## 题目：[点击打开链接][Link 1] ##

## 题意：用k种颜色对n个珠子构成的环上色，旋转、翻转后相同的只算一种，求不等价的着色方案数。 ##

## 分析：polay定理模板题。 ##

![70][]

## 代码： ##

#pragma GCC optimize(2)
    #pragma GCC optimize(3)
    #pragma GCC optimize(4)
    #pragma comment(linker, "/STACK:102400000,102400000")
    ///#include<unordered_map>
    ///#include<unordered_set>
    #include<algorithm>
    #include<iostream>
    #include<fstream>
    #include<complex>
    #include<cstdlib>
    #include<cstring>
    #include<cassert>
    #include<iomanip>
    #include<string>
    #include<cstdio>
    #include<bitset>
    #include<vector>
    #include<cctype>
    #include<cmath>
    #include<ctime>
    #include<stack>
    #include<queue>
    #include<deque>
    #include<list>
    #include<set>
    #include<map>
    using namespace std;
    #define pt(a) cout<<a<<endl
    #define debug test
    #define mst(ss,b) memset((ss),(b),sizeof(ss))
    #define rep(i,a,n) for (int i=a;i<=n;i++)
    #define per(i,a,n) for (int i=n-1;i>=a;i--)
    #define all(x) (x).begin(),(x).end()
    #define fi first
    #define se second
    #define SZ(x) ((int)(x).size())
    #define ll long long
    #define ull unsigned long long
    #define pb push_back
    #define mp make_pair
    #define inf 0x3f3f3f3f
    #define eps 1e-10
    #define PI acos(-1.0)
    typedef pair<int,int> PII;
    const ll mod = 1e9+7;
    const int N = 1e6+10;
    
    ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}
    ///ll qp(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
    ll qp(ll a,ll b) {ll res=1; for(;b;b>>=1){if(b&1)res=res*a;a=a*a;}return res;}
    int to[4][2]={
       {-1,0},{1,0},{0,-1},{0,1}};
    
    int n,m;
    
    int main() {
        ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
        while (cin>>m>>n,n||m) {
            ///旋转的情况
            ll ans = 0;
            rep(i,0,n-1) ans = ans + qp((ll) m, (ll) gcd(n, i));      ///注意这里不用乘以n
            ///翻转的情况
            if (n & 1) ans += n * qp((ll) m, (ll) n / 2 + 1);            ///若n为奇数,以一个顶点和另外一条边中点的连线为对称轴
            else ans += n / 2 * (qp((ll) m, (ll) n / 2) + qp((ll) m, (ll) n / 2 + 1)); ///n为偶数时，以两个顶点连线为对称轴  和  以两个顶点之间的连线为对称轴的情况
            cout<<ans/2/n<<endl; ///(2*n)==n+n(n为奇数)或者是n+(n/2+n/2)
        }
        return 0;
    }

[Link 1]: https://vjudge.net/problem/POJ-2409
[70]: /images/20220515/a8d1061a5a7c406fa92bf4333fb00e8b.png