cdoj802-Just a Line

http://acm.uestc.edu.cn/#/problem/show/802

Just a Line

Time Limit: 3000/1000MS (Java/Others)     Memory Limit: 65535/65535KB (Java/Others)

Submit Status

There are N points on a plane, among them N−1 points will form a line, your task is to find the point that is not on the line.

Input

The first line contains a single number N, the number of points. (4≤N≤50000)

Then come N lines each with two numbers (xi,yi), giving the position of the points. The points are given in integers. No two points' positions are the same. (−109≤xi,yi≤109)

Output

Output the position of the point that is not on the line.

Sample input and output

Sample Input	Sample Output
5 0 0 1 1 3 4 2 2 4 4	3 4

题目很简单，我一直wa的原因在于两点：%g与%.0f没有用好，遇到double型等的整数，别用%g，用%d或%.0f(我一直不晓得%g哪里错了，难道说有的整数强转后还能转出几位小数出来不成)；另一点是eps，精度之前调为1e-10竟然精度还不够，以后就直接上-20.

代码1：

 #include <fstream>
 #include <iostream>
 #include <algorithm>
 #include <cstdio>
 #include <cstring>
 #include <cmath>
 #include <cstdlib>

 using namespace std;

 #define PI acos(-1.0)
 #define EPS 1e-20
 #define lll __int64
 #define ll long long
 #define INF 0x7fffffff

 double a[][];
 int n;

 inline double K(int i,int j);//斜率
 inline bool F(double i,double j);

 int main(){
     //freopen("D:\\input.in","r",stdin);
     //freopen("D:\\output.out","w",stdout);
     scanf("%d",&n);
     n-=;
     for(int i=;i<;i++)    scanf("%lf %lf",&a[i][],&a[i][]);
     double k01=K(,);
     double k02=K(,);
     double k03=K(,);
     double k12=K(,);
     double k13=K(,);
     double k23=K(,);
     if(F(k01,k02)&&F(k01,k03)&&F(k03,k02))  printf("%.0f %.0f\n",a[][],a[][]);
     else if(F(k01,k12)&&F(k01,k13)&&F(k13,k12))  printf("%.0f %.0f\n",a[][],a[][]);
     else if(F(k12,k02)&&F(k12,k23)&&F(k23,k02))  printf("%.0f %.0f\n",a[][],a[][]);
     else if(F(k03,k13)&&F(k23,k03)&&F(k23,k13))  printf("%.0f %.0f\n",a[][],a[][]);
     else{
         double k04,k14,k24;
         for(int i=;i<n;i++){
             scanf("%lf %lf",&a[][],&a[][]);
             k04=K(,);
             k14=K(,);
             k24=K(,);
             if(F(k04,k14)&&F(k04,k24)&&F(k14,k24)){
                 printf("%.0f %.0f\n",a[][],a[][]);
                 break;
             }
         }
     }
     return ;
 }
 inline double K(int i,int j){
     if(a[i][]==a[j][])    return (double)INF;
     else    return (a[i][]-a[j][])/(a[i][]-a[j][]);
 }
 inline bool F(double i,double j){
     return fabs(i-j)>EPS;
 }

代码2：

 #include <fstream>
 #include <iostream>
 #include <algorithm>
 #include <cstdio>
 #include <cstring>
 #include <cmath>
 #include <cstdlib>

 using namespace std;

 #define PI acos(-1.0)
 #define EPS 1e-20
 #define lll __int64
 #define ll long long
 #define INF 0x7fffffff
 #define INT 2147483646

 double a[][];
 int n;

 inline double K(int i,int j);

 int main(){
     //freopen("D:\\input.in","r",stdin);
     //freopen("D:\\output.out","w",stdout);
     double kk[];
     int cnt;
     scanf("%d",&n);
     for(int i=;i<n;i++)    scanf("%lf%lf",&a[i][],&a[i][]);
     for(int i=;i<n;i++){
         cnt=;
         for(int j=;j<;j++){
             if(i==j)    continue;
             kk[cnt++]=K(i,j);
         }
         if(fabs(kk[]-kk[])>EPS&&fabs(kk[]-kk[])>EPS&&fabs(kk[]-kk[])>EPS){
             printf("%.0f %.0f\n",a[i][],a[i][]);
             break;
         }
     }
     return ;
 }
 inline double K(int i,int j){
     if(a[i][]==a[j][])    return (double)INT;
     else    return (a[i][]-a[j][])/(a[i][]-a[j][]);
 }