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：

 1 #include <fstream>

 2 #include <iostream>

 3 #include <algorithm>

 4 #include <cstdio>

 5 #include <cstring>

 6 #include <cmath>

 7 #include <cstdlib>

 8 

 9 using namespace std;

10 

11 #define PI acos(-1.0)

12 #define EPS 1e-20

13 #define lll __int64

14 #define ll long long

15 #define INF 0x7fffffff

16 

17 double a[10][2];

18 int n;

19 

20 inline double K(int i,int j);//斜率

21 inline bool F(double i,double j);

22 

23 int main(){

24     //freopen("D:\\input.in","r",stdin);

25     //freopen("D:\\output.out","w",stdout);

26     scanf("%d",&n);

27     n-=4;

28     for(int i=0;i<4;i++)    scanf("%lf %lf",&a[i][0],&a[i][1]);

29     double k01=K(0,1);

30     double k02=K(0,2);

31     double k03=K(0,3);

32     double k12=K(2,1);

33     double k13=K(3,1);

34     double k23=K(2,3);

35     if(F(k01,k02)&&F(k01,k03)&&F(k03,k02))  printf("%.0f %.0f\n",a[0][0],a[0][1]);

36     else if(F(k01,k12)&&F(k01,k13)&&F(k13,k12))  printf("%.0f %.0f\n",a[1][0],a[1][1]);

37     else if(F(k12,k02)&&F(k12,k23)&&F(k23,k02))  printf("%.0f %.0f\n",a[2][0],a[2][1]);

38     else if(F(k03,k13)&&F(k23,k03)&&F(k23,k13))  printf("%.0f %.0f\n",a[3][0],a[3][1]);

39     else{

40         double k04,k14,k24;

41         for(int i=0;i<n;i++){

42             scanf("%lf %lf",&a[4][0],&a[4][1]);

43             k04=K(0,4);

44             k14=K(1,4);

45             k24=K(2,4);

46             if(F(k04,k14)&&F(k04,k24)&&F(k14,k24)){

47                 printf("%.0f %.0f\n",a[4][0],a[4][1]);

48                 break;

49             }

50         }

51     }

52     return 0;

53 }

54 inline double K(int i,int j){

55     if(a[i][0]==a[j][0])    return (double)INF;

56     else    return (a[i][1]-a[j][1])/(a[i][0]-a[j][0]);

57 }

58 inline bool F(double i,double j){

59     return fabs(i-j)>EPS;

60 }

View Code

代码2：

 1 #include <fstream>

 2 #include <iostream>

 3 #include <algorithm>

 4 #include <cstdio>

 5 #include <cstring>

 6 #include <cmath>

 7 #include <cstdlib>

 8 

 9 using namespace std;

10 

11 #define PI acos(-1.0)

12 #define EPS 1e-20

13 #define lll __int64

14 #define ll long long

15 #define INF 0x7fffffff

16 #define INT 2147483646

17 

18 double a[50005][2];

19 int n;

20 

21 inline double K(int i,int j);

22 

23 int main(){

24     //freopen("D:\\input.in","r",stdin);

25     //freopen("D:\\output.out","w",stdout);

26     double kk[5];

27     int cnt;

28     scanf("%d",&n);

29     for(int i=0;i<n;i++)    scanf("%lf%lf",&a[i][0],&a[i][1]);

30     for(int i=0;i<n;i++){

31         cnt=0;

32         for(int j=0;j<4;j++){

33             if(i==j)    continue;

34             kk[cnt++]=K(i,j);

35         }

36         if(fabs(kk[0]-kk[1])>EPS&&fabs(kk[0]-kk[2])>EPS&&fabs(kk[1]-kk[2])>EPS){

37             printf("%.0f %.0f\n",a[i][0],a[i][1]);

38             break;

39         }

40     }

41     return 0;

42 }

43 inline double K(int i,int j){

44     if(a[i][0]==a[j][0])    return (double)INT;

45     else    return (a[i][1]-a[j][1])/(a[i][0]-a[j][0]);

46 }

View Code