HDU3441：Rotation（Polya）

传送门
题意：有一个A∗A的正方形，拆成A∗A个1∗1的小正方形，然后组成k个B∗B的正方形，而且剩下一个小正方形，也就是A∗A=K∗B∗B+1。中间小小正方形连到K个B∗B正方形的形状有多少种，有C种颜色，而且旋转视为等价。

题解：
Polya找出每个B对应的合法染色数，再找出在这个B下的所有合法解。

#include<iostream>#include<cstdio>#include<algorithm>#include<vector>using namespace std;struct IO{    streambuf *ib,*ob;    inline void init()    {        ios::sync_with_stdio(false);        cin.tie(NULL);cout.tie(NULL);        ib=cin.rdbuf();ob=cout.rdbuf();    }    inline int read()    {        static char ch;        static int i,f;        i=0,f=1;ch=ib->sbumpc();        while(!isdigit(ch)){if(ch=='-')f=-1;ch=ib->sbumpc();}        while(isdigit(ch)){i=((i+(i<<2))<<1)+ch-'0';ch=ib->sbumpc();}        return (f==1)?i:-i;    }    inline void W(int x)    {        static int buf[50];        if(!x){ob->sputc('0');return;}        if(x<0){ob->sputc('-');x=-x;}        while(x)buf[++buf[0]]=x%10,x/=10;        while(buf[0]){ob->sputc(buf[buf[0]--]+'0');}        return;    }    inline void W(string x)    {        int len=x.length();        for(int i=0;i<len;i++)ob->sputc(x[i]);    }}io;typedef pair<int,int> pii;int A,C,ans;const int Mod=1e9+7;vector<int>factory;vector<pii>fac2;vector<pii>fac3;inline void fac(int a){    for(int d=2;d*d<=a;d++)    {        if(a%d)continue;        while(!(a%d))factory.push_back(d),a/=d;    }    if(a!=1)factory.push_back(a);}inline void getfac(int a){    factory.clear();    fac(a+1),fac(a-1);    sort(factory.begin(),factory.end());    fac2.clear();fac3.clear();    fac2.push_back(make_pair(factory[0],1));    for(int e=1;e<factory.size();e++)    (factory[e]!=factory[e-1])?(fac2.push_back(make_pair(factory[e],1))):((void)fac2[fac2.size()-1].second++);}inline int power(int a,long long b){    int res=1;    for(;b;b>>=1,a=1ll*a*a%Mod)(b&1)?(res=1ll*res*a%Mod):0;    return res;}inline void dfs2(int pos,long long sze,long long phi,int &res,long long all,int k){    if(pos==fac3.size())    {        (res+=1ll*phi%Mod*power(k,1ll*all/sze)%Mod)%=Mod;        return;    }    dfs2(pos+1,sze,phi,res,all,k);    for(int i=1;i<=fac3[pos].second;i++)    {        sze*=fac3[pos].first;        (i==1)?(phi*=(fac3[pos].first-1)):(phi*=(fac3[pos].first));        dfs2(pos+1,sze,phi%Mod,res,all,k);    }}inline int calcsum(long long n,int k){    int res=0;    dfs2(0,1,1,res,n,k);    return res;}inline int calckind(int a){    int res=0;    (res+=1ll*power(C,1ll*a*a))%=Mod;    (res+=2ll*power(C,1ll*a*a/4+(a&1))%Mod)%=Mod;    (res+=1ll*power(C,1ll*a*a/2+(a&1)))%=Mod;    return 1ll*res*power(4,Mod-2)%Mod;}inline void calc(int b){    long long res=(1ll*A*A-1)/b/b;    (ans+=1ll*calcsum(res,calckind(b))*C%Mod*power(res%Mod,Mod-2)%Mod)%=Mod;}inline void dfs(int b,int pos){    if(pos==fac2.size()){calc(b);}    else    {        fac3.push_back(make_pair(fac2[pos].first,fac2[pos].second));        dfs(b,pos+1);        for(int i=2;i<=fac2[pos].second;i+=2)        {            b=b*fac2[pos].first;            fac3[fac3.size()-1].second-=2;            dfs(b,pos+1);        }        fac3.erase(--fac3.end());    }}int main(){    io.init();int T=io.read();    for(int tt=1;tt<=T;tt++)    {        A=io.read(),C=io.read();ans=0;        io.W("Case ");io.W(tt);io.W(": ");        if(A==1){io.W(C);io.ob->sputc('\n');continue;}        getfac(A);        dfs(1,0);        io.W(ans);io.ob->sputc('\n');    }}