nyoj 1241 Distribution 第八届河南省程序设计大赛

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原创最新推荐文章于 2026-03-25 00:55:18 发布 803 阅读 AI 写同款· GEO 优化 ›

机器学习小组知识点23：F分布（F Distribution）-R语言代码 F分布的用途：用于方差分析、协方差分析和回归分析等。定义一个FF-分布的随机变量是两个卡方分布变量的比率： U1/d1U2/d2=U1/U2d1/d2{\displaystyle {\frac {U_{1}/d_{1}}{U_{2}/d_{2}}}={\frac {U_{1}/U_{2}}{d_{1}/d_{2}}}} 其中：U1和U2呈卡方分布，它们的自由度（degree of freedo 阅读详情

Distribution

时间限制： 1000 ms | 内存限制： 65535 KB

难度： 1

描述

One day , Wang and Dong in the Dubai desert expedition, discovered an ancient castle. Fortunately, they found a map of the castle.The map marks the location of treasures.

They agreed to distribute the treasures according to the following rules:

Wang draws a horizontal line on the map and then Dong draws a vertical one so that the map is divided into 4 parts, as show below.

Wang will save the treasures in I and III ,while those situated in II and IV will be taken away by Dong. Wang first draw a horizontal line, Dong after the draw a vertical line.

They drew several pairs of lines. For each pair, Wang wants to know the difference between their treasures.

It's guaranteed that all the reasures will lie on neither of the lines drew by them.

输入

the first line contains two integers N and M, where N is the number of treasures on the map and M indicates how many times they are going to draw the lines. The 2nd to (N+1)-th lines Xi, Yi contain the co-ordinates of the treasures and the last M lines consist of the M pairs integers (X, Y) which means that the two splitting lines intersect at point (X, Y).
( 0 < N, M ≤ 100, 0 ≤ Xi, Yi, X,Y ≤ 1000 )

输出

Output contains M lines , a single line with a integer , the difference described above

样例输入

样例输出

-6

来源

第八届河南省程序设计大赛

 
#include<stdio.h>
struct node
{
 int x,y;
}num[1005];
int main()
{
 int m,n;
 int i,j;
 int sumw,sumk;
 scanf("%d %d",&m,&n);
 for(i=0;i<m;i++)
 {
  scanf("%d %d",&num[i].x,&num[i].y);
 }
 int a,b;
 while(n--)
 {
  sumw=0,sumk=0;
  scanf("%d %d",&a,&b);
  for(i=0;i<m;i++)
  {
   if((num[i].x>a&&num[i].y>b)||(num[i].x<a&&num[i].y<b)) sumw++;
   if((num[i].x<a&&num[i].y>b)||(num[i].x>a&&num[i].y<b)) sumk++;
  }
  printf("%d\n",sumw-sumk);
 }
 return 0;
}