数据不大，枚举哪个式子错了，对剩下的d+2个式子随意选d+1个高斯消元，然后代入剩下的式子检查是否正确，正确就是那一个式子错了
#include<bits/stdc++.h>
#define il inline
#define vd void
typedef long long ll;
il int gi(){
int x=0,f=1;
char ch=getchar();
while(!isdigit(ch)){
if(ch=='-')f=-1;
ch=getchar();
}
while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
return x*f;
}
#define eps 1e-3
double s[11][11];
double S[11];
il vd work(int n){
for(int i=1;i<=n;++i){
int t=0;
for(int j=i;j<=n;++j)if(fabs(s[i][j])>eps)t=j;
std::swap(s[t],s[i]);
for(int j=i+1;j<=n;++j){
if(fabs(s[j][i])<eps)continue;
for(int k=n+1;k>=i;--k)s[j][k]=s[j][k]*s[i][i]/s[j][i]-s[i][k];
}
}
for(int i=n;i;--i){
s[i][n+1]/=s[i][i];
for(int j=1;j<i;++j)s[j][n+1]-=s[j][i]*s[i][n+1];
}
}
int main(){
#ifndef ONLINE_JUDGE
freopen("4689.in","r",stdin);
freopen("4689.out","w",stdout);
#endif
int d;
while(scanf("%d",&d),d){
for(int i=1;i<=d+3;++i) scanf("%lf",&S[i]);
for(int i=1;i<=d+3;++i){
int j=i==1?2:1,cnt=0;
for(int k=2;k<=d+3;++k)
if(k!=i&&k!=j){
++cnt;
for(int l=1;l<=d+1;++l)s[cnt][l]=pow(k-1,l-1);
s[cnt][d+2]=S[k];
}
work(d+1);
double sum=0;
for(int k=1;k<=d+1;++k)sum+=s[k][d+2]*pow(j-1,k-1);
//printf("%d %.6lf %.6lf
",i,sum,S[j]);
if(fabs(sum-S[j])<eps)printf("%d
",i-1);
}
}
return 0;
}