acm题,希望给出代码或思路.(效率高的)这个题是卡时间的Description Staginner ,the boss of A0A(Avoid 0 AC) Company,gives you a simple task,that is to calculate how many different points are there in the group of N points.Input Th
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acm题,希望给出代码或思路.(效率高的)这个题是卡时间的Description Staginner ,the boss of A0A(Avoid 0 AC) Company,gives you a simple task,that is to calculate how many different points are there in the group of N points.Input Th
acm题,希望给出代码或思路.(效率高的)这个题是卡时间的
Description
Staginner ,the boss of A0A(Avoid 0 AC) Company,gives you a simple task,that is to calculate how many different points are there in the group of N points.
Input
There are several test cases.
The first line of each case contains a integer N(1
acm题,希望给出代码或思路.(效率高的)这个题是卡时间的Description Staginner ,the boss of A0A(Avoid 0 AC) Company,gives you a simple task,that is to calculate how many different points are there in the group of N points.Input Th
这题其实非常的简单
题目中说了座标不大于1000的非负整数
可以开一个ext[1001][10001]的bool数组
每次输入一个点的时候ext[x][y]=true
最后统计一下哪些点是true的就行了,或者可以对所有的点进行排序,然后统计.不过这样的话编程有些麻烦