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[10][2];
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-=4;
    for(int i=0;i<4;i++)    scanf("%lf %lf",&a[i][0],&a[i][1]);
    double k01=K(0,1);
    double k02=K(0,2);
    double k03=K(0,3);
    double k12=K(2,1);
    double k13=K(3,1);
    double k23=K(2,3);
    if(F(k01,k02)&&F(k01,k03)&&F(k03,k02))  printf("%.0f %.0f
",a[0][0],a[0][1]);
    else if(F(k01,k12)&&F(k01,k13)&&F(k13,k12))  printf("%.0f %.0f
",a[1][0],a[1][1]);
    else if(F(k12,k02)&&F(k12,k23)&&F(k23,k02))  printf("%.0f %.0f
",a[2][0],a[2][1]);
    else if(F(k03,k13)&&F(k23,k03)&&F(k23,k13))  printf("%.0f %.0f
",a[3][0],a[3][1]);
    else{
        double k04,k14,k24;
        for(int i=0;i<n;i++){
            scanf("%lf %lf",&a[4][0],&a[4][1]);
            k04=K(0,4);
            k14=K(1,4);
            k24=K(2,4);
            if(F(k04,k14)&&F(k04,k24)&&F(k14,k24)){
                printf("%.0f %.0f
",a[4][0],a[4][1]);
                break;
            }
        }
    }
    return 0;
}
inline double K(int i,int j){
    if(a[i][0]==a[j][0])    return (double)INF;
    else    return (a[i][1]-a[j][1])/(a[i][0]-a[j][0]);
}
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[50005][2];
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[5];
    int cnt;
    scanf("%d",&n);
    for(int i=0;i<n;i++)    scanf("%lf%lf",&a[i][0],&a[i][1]);
    for(int i=0;i<n;i++){
        cnt=0;
        for(int j=0;j<4;j++){
            if(i==j)    continue;
            kk[cnt++]=K(i,j);
        }
        if(fabs(kk[0]-kk[1])>EPS&&fabs(kk[0]-kk[2])>EPS&&fabs(kk[1]-kk[2])>EPS){
            printf("%.0f %.0f
",a[i][0],a[i][1]);
            break;
        }
    }
    return 0;
}
inline double K(int i,int j){
    if(a[i][0]==a[j][0])    return (double)INT;
    else    return (a[i][1]-a[j][1])/(a[i][0]-a[j][0]);
}