POJ 2065 SETI [高斯消元同余]

题意自己看，反正是裸题...

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普通高斯消元全换成模意义下行了

模模模！

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
const int N=;
inline int read(){
    char c=getchar();int x=,f=;
    while(c<''||c>''){if(c=='-')f=-;c=getchar();}
    while(c>=''&&c<=''){x=x*+c-'';c=getchar();}
    return x*f;
}
typedef int Matrix[N][N];
int n,P;
Matrix a;
char s[N];
inline int Pow(int a,int b){
    int re=;
    for(;b;b>>=,a=a*a%P)
        if(b&) re=re*a%P;
    return re;
}
inline int Inv(int a){return Pow(a,P-);}
void ini(){memset(a,,sizeof(a));}
void Gauss(){
    for(int i=;i<=n;i++){
        int j=i;
        while(j<=n&&!a[j][i]) j++;
        if(j!=i) for(int k=;k<=n+;k++) swap(a[i][k],a[j][k]);
        int inv=Inv(a[i][i]);
        for(int j=i+;j<=n;j++){
            int t=a[j][i]*inv%P;
            for(int k=i;k<=n+;k++) a[j][k]=(a[j][k]-t*a[i][k]%P+P)%P;
        }
    }
    for(int i=n;i>=;i--){
        for(int j=n;j>i;j--) a[i][n+]=(a[i][n+]-a[j][n+]*a[i][j]%P+P)%P;
        a[i][n+]=a[i][n+]*Inv(a[i][i])%P;
    }
}
void buildEquation(){
    ini();
    for(int i=;i<=n;i++){
        for(int j=;j<=n;j++) a[i][j]=Pow(i,j-);//,printf("%d%c",a[i][j],j==n?'\n':' ');
        a[i][n+]=s[i]=='*'?:s[i]-'a'+;
    }
}
int main(){
    freopen("in","r",stdin);
    int T=read();
    while(T--){
        P=read();
        scanf("%s",s+);
        n=strlen(s+);
        buildEquation();
        Gauss();
        for(int i=;i<=n;i++) printf("%d ",a[i][n+]);
        puts("");
    }
}