枚举1个不选,选的n+2个里随便选n+1个高斯消元出多项式,然后看与剩下的那个是否相等,相等的话说明没选那个是错的
精度不用太高
#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<cstring>
#include<ctime>
#include<cmath>
#include<algorithm>
#include<iomanip>
#include<queue>
#include<map>
#include<bitset>
#include<stack>
#include<vector>
#include<set>
using namespace std;
#define MAXN 1010
#define MAXM 1010
#define INF 1000000000
#define MOD 1000000007
#define ll long long
#define eps 5e-6
int n,N;
double f[10];
double A[10];
double a[10][10];
int TOT;
bool gs(){
int i,j,k;
for(i=0;i<=n;i++){
if(fabs(a[i][i])<eps){
for(j=i+1;j<=n;j++){
if(fabs(a[j][i])>eps){
for(k=0;k<=n+1;k++){
swap(a[i][k],a[j][k]);
}
break;
}
}
}
if(fabs(a[i][i])>eps){
for(j=0;j<=n;j++){
if(i!=j){
double t=a[j][i]/a[i][i];
for(k=0;k<=n+1;k++){
a[j][k]-=t*a[i][k];
}
}
}
}else{
return 0;
}
}
return 1;
}
int main(){
int i,j,k,l;
while(scanf("%d",&n)){
if(!n){
break;
}
for(i=1;i<=n+3;i++){
scanf("%lf",&f[i]);
}
for(i=1;i<=n+3;i++){
int tot=0;
for(k=1;tot<=n+1;k++){
if(k!=i){
for(l=0;l<=n;l++){
a[tot][l]=pow(1.0*k,l);
}
a[tot][n+1]=f[k];
tot++;
}
}
if(!gs()){
continue ;
}
bool flag=1;
for(k=0;k<=n;k++){
if(fabs(a[k][k])<eps){
flag=0;
break;
}
A[k]=a[k][n+1]/a[k][k];
}
for(k=n+3;k;k--){
if(k!=i){
break;
}
}
double t=0;
for(l=0;l<=n;l++){
t+=A[l]*pow(1.0*k,l);
}
if(fabs(t-f[k])<0.5){
printf("%d\n",i-1);
break;
}
}
}
return 0;
}
/*
2
1.0
4.0
12.0
16.0
25.0
1
-30.5893962764
5.76397083962
39.3853798058
74.3727663177
4
42.4715310246
79.5420238202
28.0282396675
-30.3627807522
-49.8363481393
-25.5101480106
7.58575761381
5
-21.9161699038
-48.469304271
-24.3188578417
-2.35085940324
-9.70239202086
-47.2709510623
-93.5066246072
-82.5073836498
0
*/