UVALive5379 UVA270 Lining Up【输入输出＋水题】

“How am I ever going to solve this problem?” said the pilot.

  Indeed, the pilot was not facing an easy task. She had to drop packages at specific points scatteredin a dangerous area. Furthermore, the pilot could only fly over the area once in a straight line, and shehad to fly over as many points as possible. All points were given by means of integer coordinates in atwo-dimensional space. The pilot wanted to know the largest number of points from the given set thatall lie on one line. Can you write a program that calculates this number?

  Your program has to be efficient!

Input

The input begins with a single positive integer on a line by itself indicating the number of the casesfollowing, each of them as described below. This line is followed by a blank line, and there is also ablank line between two consecutive inputs.

  The input consists of N pairs of integers, where 1 < N < 700. Each pair of integers is separated byone blank and ended by a new-line character. The list of pairs is ended with an end-of-file character.No pair will occur twice.

Output

For each test case, the output must follow the description below. The outputs of two consecutive caseswill be separated by a blank line.

  The output consists of one integer representing the largest number of points that all lie on one line.

Sample Input

1

1 1

2 2

3 3

9 10

10 11

Sample Output

3

Regionals 1994 >> North America - East Central NA

问题链接：UVALive5379 UVA270 Lining Up。

题意简述：

　　输入n，输入n个整数对，即n个坐标点，问最多共线点数是多少。

问题分析：

  用暴力法解决这个问题，好在计算规模不算大。

程序说明：

  程序中，判断共线时，使用的是乘法，没有用除法，可以保证精确的计算结果。

  特别需要说明的是，这个问题虽然与参考链接的题是相同的问题，但是数据输入格式不一样，需要特别处理。程序中编写了函数mygetline()实现读入一行，用以代替C语言的库函数gets()（新的C函数库标准中，已将函数gets()剔除，因为该函数容易引起问题）。函数mygetline()中，需要考虑EOF的情形，这是程序员容易忽略的地方。

　　另外，程序中还使用了函数sscanf()。

参考链接：POJ1118 HDU1432 Lining Up

　

AC的C语言程序如下：

/* UVALive5379 UVA270 Lining Up */

#include 

#define MAXN 700

struct {
    int x, y;
} p[MAXN];      /* point */

void mygetline(char *pc)
{
    char c;

    while((c=getchar()) != '\n' && c !=EOF)
        *pc++ = c;
    *pc = '\0';
}

int main(void)
{
    int t, n, ans, max, i, j, k;
    char s[128];

    scanf("%d", &t);
    getchar();
    getchar();
    while(t--) {
        n = 0;
        mygetline(s);
        while(s[0] != '\0') {
            sscanf(s, "%d%d", &p[n].x, &p[n].y);
            n++;
            mygetline(s);
        }

        ans = 2;
        for(i=0; i ans)
                    ans = max;
            }

        printf("%d\n", ans);

        if(t)
            printf("\n");
    }

    return 0;
}