sscanf函数的用法

最新推荐文章于 2025-05-11 09:06:44 发布

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最新推荐文章于 2025-05-11 09:06:44 发布

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B - Lining Up

Time Limit:3000MS Memory Limit:0KB 64bit IO Format:%lld & %llu

Submit Status

Description

Lining Up

``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 scattered in a dangerous area. Furthermore, the pilot could only fly over the area once in a straight line, and she had to fly over as many points as possible. All points were given by means of integer coordinates in a two-dimensional space. The pilot wanted to know the largest number of points from the given set that all 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 cases following, each of them as described below. This line is followed by a blank line, and there is also a blank line between two consecutive inputs.
The input consists of N pairs of integers, where 1 < N < 700. Each pair of integers is separated by one 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 cases will 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

Sample Output

来源： <http://acm.hust.edu.cn/vjudge/contest/view.action?cid=99978>

题意：

给出很多点的坐标，算出在一条直线的点最多的个数。

不过输入的时候，注意要用字符串的输入来判断输入是否停止。

代码：

     
      
       #include<iostream>
      
      
       #include<cstdio>
      
      
       #include<cstring>
      
      
       #include<queue>
      
      
       #include <stack>
      
      
       #include <string>
      
      
       #include <set>
      
      
       #include <map>
      
      
       #include <cstdlib>
      
      
       #include <cmath>
      
      
       #include <algorithm>
      
      
       #define MAXX 10000
      
      
        
      
      
       struct point
      
      
       {
      
      
           int x, y;
      
      
           bool operator < (const point &a) const
      
      
           {
      
      
               if (a.x == x)
      
      
                   return y<a.y;
      
      
               return x<a.x;
      
      
           }
      
      
       }p[MAXX];
      
      
       using namespace std;
      
      
       double k[MAXX];
      
      
       int main()
      
      
       {
      
      
           #ifdef LOCAL
      
      
               freopen("in.txt", "r", stdin);
      
      
               //freopen("out.txt", "w", stdout);
      
      
           #endif // LOCAL
      
      
           int T;
      
      
           scanf("%d", &T);
      
      
           cin.get();
      
      
           cin.get();
      
      
           while (T--)
      
      
           {
      
      
               int n=0, x, y;
      
      
               char str[100];
      
      
               while(gets(str))
      
      
               {
      
      
                   if (str[0]=='\0') break;
      
      
                   sscanf(str,"%d%d",&p[n].x,&p[n].y);
      
      
                   n++;
      
      
               }
      
      
               int ans=0;
      
      
               sort(p, p+n);
      
      
               for (int i=0; i<n-1; i++)
      
      
               {
      
      
                   for (int j=i+1; j<n; j++)
      
      
                   {
      
      
                       int maxx = 0;
      
      
                       int A=p[i].y-p[j].y;
      
      
                       int B=p[i].x-p[j].x;
      
      
                       int C=-p[i].x*A+p[i].y*B;
      
      
                       for (int k=0; k<n; k++)
      
      
                           if ((-B*p[k].y+A*p[k].x+C) == 0)
      
      
                           maxx++;
      
      
                       ans = max(ans, maxx);
      
      
                   }
      
      
               }
      
      
               cout<<ans<<endl;
      
      
               if(T)
      
      
                   cout<<endl;
      
      
           }
      
      
           return 0;
      
      
       }

方法：

先算出一条直线的一般式的A、B、C的值，然后在一个一个的点去判断是否在直线上

相关知识：

sscanf函数用法，是从字符串里读出与指定格式相符的数据。

1. 常见用法。
　　char buf[512] = ;
　　sscanf("123456 ", "%s", buf);
　　printf("%s\n", buf);
　　结果为：123456
　　2. 取指定长度的字符串。如在下例中，取最大长度为4字节的字符串。
　　sscanf("123456 ", "%4s", buf);
　　printf("%s\n", buf);
　　结果为：1234
　　3. 取到指定字符为止的字符串。如在下例中，取遇到空格为止字符串。
　　sscanf("123456 abcdedf", "%[^ ]", buf);
　　printf("%s\n", buf);

结果为：123456
　　4. 取仅包含指定字符集的字符串。如在下例中，取仅包含1到9和小写字母的字符串。
　　sscanf("123456abcdedfBCDEF", "%[1-9a-z]", buf);
　　printf("%s\n", buf);
　　结果为：123456abcdedf
　　5. 取到指定字符集为止的字符串。如在下例中，取遇到大写字母为止的字符串。
　　sscanf("123456abcdedfBCDEF", "%[^A-Z]", buf);
　　printf("%s\n", buf);
　　结果为：123456abcdedf
　　6、给定一个字符串iios/12DDWDFF@122，获取 / 和 @ 之间的字符串，先将 "iios/"过滤掉，再将非'@'的一串内容送到buf中
　　sscanf("iios/12DDWDFF@122", "%*[^/]/%[^@]", buf);
　　printf("%s\n", buf);

　7、给定一个字符串““hello, world”，仅保留world。（注意：“，”之后有一空格）
　　sscanf(“hello, world”, "%*s%s", buf);
　　printf("%s\n", buf);
　　结果为：world
　　%*s表示第一个匹配到的%s被过滤掉，即hello被过滤了
　　如果没有空格则结果为NULL。
　　sscanf的功能很类似于正则表达式, 但却没有正则表达式强大,所以如果对于比较复杂的字符串处理,建议使用正则表达式.
　　//-------------------------------------------------------
　　sscanf,表示从字符串中格式化输入
　　上面表示从str中，输入数字给x，就是32700
　　久以前，我以为c没有自己的split string函数，后来我发现了sscanf；一直以来，我以为sscanf只能以空格来界定字符串，现在我发现我错了。
　　sscanf是一个运行时函数，原形很简单：

int sscanf(
　　const char *buffer,
　　const char *format [,
　　argument ] ...
　　);
　　它强大的功能体现在对format的支持上。
　　我以前用它来分隔类似这样的字符串2006:03:18:
　　int a, b, c;
　　sscanf("2006:03:18", "%d:%d:%d", a, b, c);
　　以及2006:03:18 - 2006:04:18:
　　char sztime1[16] = "", sztime2[16] = "";
　　sscanf("2006:03:18 - 2006:04:18", "%s - %s", sztime1, sztime2);
　　但是后来，我需要处理2006:03:18-2006:04:18
　　仅仅是取消了‘-’两边的空格，却打破了%s对字符串的界定。

来源： <http://blog.163.com/xs_fantasy/blog/static/1015608712010886248407/>