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)
 

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 InputSample 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

 

posted @ 2015-03-24 13:16  jiu~  阅读(347)  评论(0)  编辑  收藏  举报