bzoj 4689: Find the Outlier

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\n",i,sum,S[j]);
			if(fabs(sum-S[j])<eps)printf("%d\n",i-1);
		}
	}
	return 0;
}

posted @ 2018-10-03 14:27  菜狗xzz  阅读(158)  评论(0编辑  收藏  举报