1 //-----------------------------------------------------------------------------
2 /*快速幂*/
3 inline int qpow(re int x,re int y,re int res=1){
4 for(;y;y>>=1,x=x*x%mod) if(y&1) res=res*x%mod; return res;
5 }
6 //-----------------------------------------------------------------------------
7 /*gcd*/
8 //gcd
9 inline int gcd(re int x,re int y){return !y?x:gcd(y,x%y);}
10 //lcm
11 inline int lcm(re int x,re int y){return x*y/gcd(x,y);}
12 //-----------------------------------------------------------------------------
13 /*exgcd*/
14 inline int exgcd(re int a,re int b,re int &x,re int &y){
15 re int ret,tmp;if(!b){x=1,y=0; return a;}
16 ret=exgcd(b,a%b,x,y),tmp=x,x=y,y=tmp-a/b*y;
17 return ret;
18 }
19 //-----------------------------------------------------------------------------
20 /*分解质因数*/
21 for(re int i=1;prime[i]*prime[i]<=x;++i){
22 while(x%prime[i]==0)
23 x/=prime[i],p[++p[0]]=prime[i];
24 }
25 if(x>1) p[++p[0]]=x;
26 //-----------------------------------------------------------------------------
27 /*线性筛*/
28 //素数
29 for(re int i=2;i<=n;++i){
30 if(!vis[i]) prime[++cnt]=i,vis[i]=1;
31 for(re int j=1;j<=cnt&&i*prime[j]<=n;++j){
32 vis[i*prime[j]]=1;
33 if(i%prime[j]==0) break;
34 }
35 }
36 //约数个数和
37 d[1]=1;
38 for(re int i=2;i<=n;++i){
39 if(!vis[i]) prime[++cnt]=i,d[i]=2,c[i]=1;
40 for(re int j=1;j<=cnt&&i*prime[j]<=n;++j){
41 vis[i*prime[j]]=1;
42 if(i%prime[j]==0){
43 c[i*prime[j]]=c[i]+1;
44 d[i*prime[j]]=d[i]/(c[i]+1)*(c[i]+2);
45 break;
46 }
47 d[i*prime[j]]=d[i]*2;
48 c[i*prime[j]]=1;
49 }
50 }
51 //约数和
52 s[1]=1;
53 for(re int i=2;i<=n;++i){
54 if(!vis[i]) prime[++cnt]=i,s[i]=i+1,c[i]=i+1;
55 for(re int j=1;j<=cnt&&i*prime[j]<=n;++j){
56 vis[i*prime[j]]=1;
57 if(i%prime[j]==0){
58 c[i*prime[j]]=c[i]*prime[j]+1;
59 s[i*prime[j]]=s[i]/c[i]*c[i*prime[j]];
60 break;
61 }
62 d[i*prime[j]]=d[i]*(prime[j]+1);
63 c[i*prime[j]]=prime[j]+1;
64 }
65 }
66 //莫比乌斯函数
67 mu[1]=1;
68 for(re int i=2;i<=n;++i){
69 if(!vis[i]) prime[++cnt]=i,mu[i]=-1;
70 for(re int j=1;j<=cnt&&i*prime[j]<=n;++j){
71 vis[i*prime[j]]=1;
72 if(i%prime[j]==0) break;
73 mu[i*prime[j]]=-mu[i]
74 }
75 }
76 //欧拉函数
77 phi[1]=1;
78 for(re int i=2;i<=n;++i){
79 if(!vis[i]) prime[++cnt]=i,phi[i]=i-1;
80 for(re int j=1;j<=cnt&&i*prime[j]<=n;++j){
81 vis[i*prime[j]]=1;
82 if(i%prime[j]==0){
83 phi[i*prime[j]]=phi[i]*prime[j];
84 break;
85 }
86 phi[i*prime[j]]=phi[i]*phi[prime[j]];
87 }
88 }
89 //求单个欧拉函数
90 inline int phi(re int x){
91 re int dat=sqrt(x),res=x;
92 for(re int i=2;i<=dat;i++)
93 if(x%i==0){
94 res=res-res/i;
95 while(x%i==0) x/=i;
96 }
97 if(x!=1) res=res-res/x;
98 return res;
99 }
100 //-----------------------------------------------------------------------------
101 /*BSGS*/
102 #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
103 tr1::unordered_map<int,int> mp;
104 inline int qpow(re int x,re int y,re int res=1){
105 for(;y;y>>=1,x=x*x%p) if(y&1) res=res*x%p; return res;
106 }
107 scanf("%d%d%d",&p,&a,&b);
108 if(a%p==0)
109 printf("no solution\n");return 0;
110 m=ceil(sqrt(p)),flag=0;
111 now=b%p;mp[now]=0;//b*a^j 当j==0时
112 for(re int i=1;i<=m;++i)
113 now=(now*a)%p,mp[now]=i;
114 dat=qpow(a,m);now=1;
115 for(re int i=1;i<=m;++i){//枚举 (a^m)^i
116 now=(now*dat)%p;
117 if(mp[now]){
118 flag=1;
119 ans=i*m-mp[now];
120 printf("%lld\n",(ans%p+p)%p);
121 break;
122 }
123 }
124 if(!flag) printf("no solution\n");
125 //-----------------------------------------------------------------------------
126 /*CRT*/
127 //n个方程：x=a[i](mod m[i]) (0<=i<n)
128 inline void exgcd(int a,int b,int &x,int &y,int &d){
129 if(!b){x=1,y=0,d=a;}
130 else exgcd(b,a%b,y,x,d),y-=x*(a/b);
131 }
132 inline int inv(re int t,re int p){
133 re int dat,x,y;
134 exgcd(t,p,x,y,dat);
135 return dat==1?(x%p+p)%p:-1;
136 }
137 inline int China(re int x,re int *a,re int *b){
138 re int dat=1,res=0;
139 for(re int i=0;i<x;++i) dat*=b[i];
140 for(re int i=0;i<x;++i){
141 re int val=dat/b[i];
142 res=(res+val*inv(val,b[i])*a[i])%dat;
143 }
144 return (res+dat)%dat;
145 }
146 //-----------------------------------------------------------------------------
147 /*排列组合*/
148 //阶乘
149 fac[0]=1;
150 for(re int i=1;i<=n;++i)
151 fac[i]=fac[i-1]*i%mod;
152 //逆元（线性）
153 inv[0]=inv[1]=1;
154 for(re int i=2;i<=n;++i)
155 inv[i]=inv[mod%i]*(mod-mod/i)%mod;
156 //组合数
157 inline int C(re int x,re int y){
158 if(x<y) return 0;if(x==y) return 1;
159 return fac[x]*inv[y]%mod*inv[x-y]%mod;
160 }
161 //排列数
162 inline int A(re int x,re int y){
163 if(x<y) return 0;if(x==y) return fac[x];
164 return fac[x]*inv[x-y]%mod;
165 }
166 //卡特兰数
167 inline int katelan(re int x){
168 return C(x<<1,x)*inv[x+1]%mod;
169 }
170 //-----------------------------------------------------------------------------
171 /*lucas*/
172 int lucas(re int x,re int y){
173 if(!y) return 1;
174 return C(x%mod,y%mod)*lucas(x/mod,y/mod)%mod;
175 }
176 //-----------------------------------------------------------------------------
177 /*高斯消元*/
178 inline void work(){
179 for(re int i=1;i<=n;++i){
180 re int now=i;
181 for(re int j=i+1;j<=n;++j)
182 if(fabs(a[j][i]>a[now][i])) now=j;
183 if(now!=i){
184 for(re int j=i;j<=n+1;++j)
185 swap(a[i][j],a[now][j]);
186 }
187 if(a[i][i]=0){flag=1; return;}
188 for(re int j=i+1;j<=n+1;++j)
189 a[i][j]=a[i][j]/a[i][i];
190 a[i][i]=1;
191 for(re int j=i+1;j<=n;++j){
192 for(re int k=i+1;k<=n+1;++k)
193 a[j][k]-=a[i][k]*a[j][i];
194 a[j][i]=0;
195 }
196 }
197 for(re int i=n;i;--i)
198 for(re int j=i+1;j<=n;++j)
199 a[i][n+1]-=a[i][j]*a[j][n+1];
200 }
201 work();if(flag)->无解
202 for(re int i=1;i<=n;i++)
203 printf("%.2lf\n",a[i][n+1]);
204 //-----------------------------------------------------------------------------
