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| const int N=405,P=1e9+7;
int n;
LL a[N][N<<1];
LL quickpow(LL a, LL b){
LL ans = 1;
while(b){
if(b & 1) ans = ans*a%P;
a = a*a%P;
b >>= 1;
}
return ans;
}
bool Gauss_Jordan(){
for(int i=1;i<=n;++i){ //枚举主元的行列
int r = i;
for(int k=i; k<=n; ++k) //找非0行
if(a[k][i]) {r=k; break;}
if(r!=i) swap(a[r],a[i]); //换行
if(!a[i][i]) return 0;
int x=quickpow(a[i][i],P-2); //求逆元
for(int k=1; k<=n; ++k){ //对角化
if(k == i) continue;
int t=a[k][i]*x%P;
for(int j=i; j<=2*n; ++j)
a[k][j]=((a[k][j]-t*a[i][j])%P+P)%P;
}
for(int j=1; j<=2*n; ++j) //除以主元
a[i][j]=(a[i][j]*x%P);
}
return 1;
}
int main(){
scanf("%d",&n);
for(int i=1; i<=n; ++i)
for(int j=1; j<=n; ++j)
scanf("%lld",&a[i][j]),a[i][i+n]=1;
if(Gauss_Jordan())
for(int i=1; i<=n; ++i){
for(int j=n+1; j<=2*n; ++j)
printf("%lld ",a[i][j]);
puts("");
}
else puts("No Solution");
return 0;
}
|
