常用数论模板

//-----------------------------------------------------------------------------
/*快速幂*/
inline int qpow(re int x,re int y,re int res=1){
    for(;y;y>>=1,x=x*x%mod) if(y&1) res=res*x%mod; return res;
}
//-----------------------------------------------------------------------------
/*gcd*/
//gcd
inline int gcd(re int x,re int y){return !y?x:gcd(y,x%y);}
//lcm
inline int lcm(re int x,re int y){return x*y/gcd(x,y);}
//-----------------------------------------------------------------------------
/*exgcd*/
inline int exgcd(re int a,re int b,re int &x,re int &y){
    re int ret,tmp;if(!b){x=1,y=0; return a;}
    ret=exgcd(b,a%b,x,y),tmp=x,x=y,y=tmp-a/b*y;
    return ret;
}
//-----------------------------------------------------------------------------
/*分解质因数*/
for(re int i=1;prime[i]*prime[i]<=x;++i){
    while(x%prime[i]==0)
    x/=prime[i],p[++p[0]]=prime[i];
}
if(x>1) p[++p[0]]=x;
//-----------------------------------------------------------------------------
/*线性筛*/
//素数
for(re int i=2;i<=n;++i){
    if(!vis[i]) prime[++cnt]=i,vis[i]=1;
    for(re int j=1;j<=cnt&&i*prime[j]<=n;++j){
        vis[i*prime[j]]=1;
        if(i%prime[j]==0) break;
    }
}
//约数个数和
d[1]=1;
for(re int i=2;i<=n;++i){
    if(!vis[i]) prime[++cnt]=i,d[i]=2,c[i]=1;
    for(re int j=1;j<=cnt&&i*prime[j]<=n;++j){
        vis[i*prime[j]]=1;
        if(i%prime[j]==0){
            c[i*prime[j]]=c[i]+1;
            d[i*prime[j]]=d[i]/(c[i]+1)*(c[i]+2);
            break;
        }
        d[i*prime[j]]=d[i]*2;
        c[i*prime[j]]=1;
    }
}
//约数和
s[1]=1;
for(re int i=2;i<=n;++i){
    if(!vis[i]) prime[++cnt]=i,s[i]=i+1,c[i]=i+1;
    for(re int j=1;j<=cnt&&i*prime[j]<=n;++j){
        vis[i*prime[j]]=1;
        if(i%prime[j]==0){
            c[i*prime[j]]=c[i]*prime[j]+1;
            s[i*prime[j]]=s[i]/c[i]*c[i*prime[j]];
            break;
        }
        d[i*prime[j]]=d[i]*(prime[j]+1);
        c[i*prime[j]]=prime[j]+1;
    }
}
//莫比乌斯函数
mu[1]=1;
for(re int i=2;i<=n;++i){
    if(!vis[i]) prime[++cnt]=i,mu[i]=-1;
    for(re int j=1;j<=cnt&&i*prime[j]<=n;++j){
        vis[i*prime[j]]=1;
        if(i%prime[j]==0) break;
        mu[i*prime[j]]=-mu[i]
    }
}
//欧拉函数
phi[1]=1;
for(re int i=2;i<=n;++i){
    if(!vis[i]) prime[++cnt]=i,phi[i]=i-1;
    for(re int j=1;j<=cnt&&i*prime[j]<=n;++j){
        vis[i*prime[j]]=1;
        if(i%prime[j]==0){
            phi[i*prime[j]]=phi[i]*prime[j]; 
            break;
        }
        phi[i*prime[j]]=phi[i]*phi[prime[j]];
    }
}
//求单个欧拉函数
inline int phi(re int x){
    re int dat=sqrt(x),res=x;
    for(re int i=2;i<=dat;i++) 
        if(x%i==0){
            res=res-res/i;
            while(x%i==0) x/=i;
        }
    if(x!=1) res=res-res/x;
    return res;
}
//-----------------------------------------------------------------------------
/*BSGS*/
#include<tr1/unordered_map>//底层为hash的map,可以减少map带来的log级别的复杂度
tr1::unordered_map<int,int> mp;
inline int qpow(re int x,re int y,re int res=1){
    for(;y;y>>=1,x=x*x%p) if(y&1) res=res*x%p; return res;
}
scanf("%d%d%d",&p,&a,&b);
if(a%p==0)
    printf("no solution\n");return 0;
m=ceil(sqrt(p)),flag=0;
now=b%p;mp[now]=0;//b*a^j 当j==0时 
for(re int i=1;i<=m;++i)
    now=(now*a)%p,mp[now]=i;
dat=qpow(a,m);now=1;
for(re int i=1;i<=m;++i){//枚举 (a^m)^i
    now=(now*dat)%p;
    if(mp[now]){
        flag=1;
        ans=i*m-mp[now];
        printf("%lld\n",(ans%p+p)%p);
        break;
    }
}
if(!flag) printf("no solution\n");
//-----------------------------------------------------------------------------
/*CRT*/
//n个方程：x=a[i](mod m[i]) (0<=i<n)
inline void exgcd(int a,int b,int &x,int &y,int &d){
    if(!b){x=1,y=0,d=a;}
    else exgcd(b,a%b,y,x,d),y-=x*(a/b);
}
inline int inv(re int t,re int p){
    re int dat,x,y;
    exgcd(t,p,x,y,dat);
    return dat==1?(x%p+p)%p:-1;
}
inline int China(re int x,re int *a,re int *b){
    re int dat=1,res=0;
    for(re int i=0;i<x;++i) dat*=b[i];
    for(re int i=0;i<x;++i){
        re int val=dat/b[i];
        res=(res+val*inv(val,b[i])*a[i])%dat;
    }
    return (res+dat)%dat;
}
//-----------------------------------------------------------------------------
/*排列组合*/
//阶乘
fac[0]=1;
for(re int i=1;i<=n;++i)
    fac[i]=fac[i-1]*i%mod;
//逆元（线性）
inv[0]=inv[1]=1;
for(re int i=2;i<=n;++i)
    inv[i]=inv[mod%i]*(mod-mod/i)%mod;
//组合数
inline int C(re int x,re int y){
    if(x<y) return 0;if(x==y) return 1;
    return fac[x]*inv[y]%mod*inv[x-y]%mod;
}
//排列数
inline int A(re int x,re int y){
    if(x<y) return 0;if(x==y) return fac[x];
    return fac[x]*inv[x-y]%mod;
}
//卡特兰数
inline int katelan(re int x){
    return C(x<<1,x)*inv[x+1]%mod;
}
//-----------------------------------------------------------------------------
/*lucas*/
int lucas(re int x,re int y){
    if(!y) return 1;
    return C(x%mod,y%mod)*lucas(x/mod,y/mod)%mod;
}
//-----------------------------------------------------------------------------
/*高斯消元*/
inline void work(){
    for(re int i=1;i<=n;++i){
        re int now=i;
        for(re int j=i+1;j<=n;++j)
            if(fabs(a[j][i]>a[now][i])) now=j;
        if(now!=i){
            for(re int j=i;j<=n+1;++j)
                swap(a[i][j],a[now][j]);
        }
        if(a[i][i]=0){flag=1; return;}
        for(re int j=i+1;j<=n+1;++j)
            a[i][j]=a[i][j]/a[i][i];
        a[i][i]=1;
        for(re int j=i+1;j<=n;++j){
            for(re int k=i+1;k<=n+1;++k)
                a[j][k]-=a[i][k]*a[j][i];
            a[j][i]=0;
        }
    }
    for(re int i=n;i;--i)
        for(re int j=i+1;j<=n;++j)
            a[i][n+1]-=a[i][j]*a[j][n+1];
}
work();if(flag)->无解
for(re int i=1;i<=n;i++)
    printf("%.2lf\n",a[i][n+1]);
//-----------------------------------------------------------------------------