題目鏈接
http://blog.csdn.net/Clove_unique/article/details/54381675
http://blog.csdn.net/u013081425/article/details/24299047
http://blog.csdn.net/lin375691011/article/details/38406737
https://www.cnblogs.com/IMGavin/p/5933037.html

/*
模意義下的高斯消元，在初等行變換時把k=tar/Anow改為k=tag*inv(Anow) 
最後求解回帶時把除法用乘逆元替代即可 
也可用擴展歐幾裏得求出一個最小的ans[i] 
 */
#include <cstdio>
 

#include <cstring>
#include <algorithm>
#define mod p
typedef long long LL;
const int N=100;

namespace Gauss
{
    int p,n,A[N][N],ans[N];
    char s[N];
    int FP(LL x,int k)
    {
        LL t=1;
        for(;k;k>>=1, x=x*x%p)
            if(k&1) t=t*x%p;
        return t;
    }
    inline
 
 int inv(int x) {return FP(x,p-2);}
    void Init()
    {
        scanf("%d%s",&p,s);
        n=strlen(s);
        for(int i=0; i<n; ++i)
        {
            A[i][0]=1;
            for(int j=1; j<n; ++j)
                A[i][j]=(i+1)*A[i][j-1]%p;
            A[i][n]= s[i]=='*'
 
?0:s[i]-'a'+1;
        }
    }
    void Solve()
    {
        Init();
        for(int j=0; j<n; ++j)
        {
            int mxrow=j;
            for(int i=j+1; i<n; ++i)
                if(A[i][j]>A[mxrow][j]) mxrow=i;
            if(mxrow!=j) std::swap(A[mxrow],A[j]);
            for(int i=j+1; i<n; ++i)
                if(A[i][j])
                {
                    int t=A[i][j]*inv(A[j][j])%mod;
                    for(int k=j; k<=n; ++k)
                        A[i][k]=((A[i][k]-t*A[j][k]%mod)%mod+mod)%mod;
                }
        }
        for(int i=n-1; ~i; --i)
        {
            for(int j=i+1; j<n; ++j)
                A[i][n]=((A[i][n]-A[i][j]*ans[j]%mod)+mod)%mod;
            ans[i]=A[i][n]*inv(A[i][i])%mod;
        }
        for(int i=0; i<n-1; ++i) printf("%d ",ans[i]);
        printf("%d\n",ans[n-1]);
    }
}

int main()
{
    int t; scanf("%d",&t);
    while(t--) Gauss::Solve();
    return 0;
}/*
3
31 aaa
37 abc
29 hello*earth
*/

POJ.2065.SETI(高斯消元 模線性方程組)