AtCoder Beginner Contest 137 F

AtCoder Beginner Contest 137 F

Number Theory ghost problem (although not particularly number theory)

I hope you view this problem before the solution has been known Fermat's Little Theorem

Fermat's little theorem by using the constructor \ (g (x) = ( xi) ^ {P-1} \)

\[x=i,g(x)=0\]

\ [X \ ne i, g (x) = 1 \]

Then we can construct

\[f(x)=\sum^{i=0}_{P-1}(-a_i*(x-i)^{P-1}+a_i)\]

For the first \ (I \) Article expression if and only if \ (a_i = 1 \ and \ x = i \) time taken to \ (1 \)

Code written in relatively strange

const int N=3100;

int P,a[N];

int po[N]={1},Inv[N]={1,1};
int b[N];
int C(int n,int m){
    if(n<0||m<0||n<m) return 0;
    return po[n]*Inv[m]%P*Inv[n-m]%P;
}

int fl=0;
int main(){
    P=rd();
    rep(i,1,P+1) po[i]=po[i-1]*i%P;
    rep(i,2,P-1) Inv[i]=(P-P/i)*Inv[P%i]%P;
    rep(i,2,P+1) Inv[i]=Inv[i]*Inv[i-1]%P;
    rep(i,0,P-1) {
        if(rd()) {
            fl=1;
            int t=1;
            drep(j,P-1,0) b[j]+=C(P-1,j)%P*t%P,t=t*(P-i)%P;
        } else b[0]++;
    }
    rep(i,0,P-1) {
        int x=(P-b[i])%P;
        x=(x%P+P)%P;
        printf("%d%c",x,(i==P-1)?'\n':' ');
    }
}