題目連接

題目描述

One day , Wang and Dong in the Dubai desertexpedition, discovered an ancient castle. Fortunately, they found a map of thecastle.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 betaken away by Dong.?Wangfirst draw a horizontal line, Dong after the draw a vertical line.

They drewseveral pairs of? lines. For each pair, Wangwants to know the difference between their treasures.

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

輸入

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 )

輸出

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

樣例輸入

10 3 29 22 17 14 18 23 3 15 6 28 30 27 4 1 26 7 8 0 11 21 2 25 5 10 19 24

樣例輸出

-6 4 4
題意大概： 在一個平面上有N個點，有M次劃分，既是以M次劃分的點為分割點，畫一個水平線和垂直線劃分為四個區域，數一數一三界限和二四界限的點的差額既是所求的結果。代碼如下： #include<stdio.h> #include<string.h> #include<iostream>using namespace std;struct node{int x;int y; }loc[105];int main(){int n,m;while(scanf("%d %d",&n,&m)!=EOF){memset(loc,0,sizeof(loc));//int i;for(int i=0; i<n; ++i){scanf("%d %d", &loc[i].x, &loc[i].y );}for(int i=0; i<m; ++i ){int a,b,s=0,s2=0,s4=0;scanf("%d %d",&a,&b);for(int j=0; j<n; ++j){if(loc[j].x>a && loc[j].y>b )//四界限的點數 s4++;else if(loc[j].x<a && loc[j].y<b)//二界限的點數 s2++;elses++;}printf("%d\n",s2+s4-s);//兩人的差額 }}return 0; }

總結

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