CodeForces 195D(脑洞)

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最新推荐文章于 2025-01-20 21:44:42 发布

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分类专栏：脑洞文章标签： CodeForces 195D

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本文介绍了一个编程任务：计算多个特殊形式的一次函数相加后得到的折线图中，不等于180度的角度数量。具体而言，需要找出所有不同交点，这些交点对应着折线图的拐点。

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问题描述:

As Valeric and Valerko were watching one of the last Euro Championship games in a sports bar, they broke a mug. Of course, the guys paid for it but the barman said that he will let them watch football in his bar only if they help his son complete a programming task. The task goes like that.

Let's consider a set of functions of the following form:

$\text{[math]}$ Let's define a sum of n functions y1(x), ..., yn(x) of the given type as functions(x) = y1(x) + ... + yn(x) for any x. It's easy to show that in this case the graph s(x)is a polyline. You are given n functions of the given type, your task is to find the number of angles that do not equal 180 degrees, in the graph s(x), that is the sum of the given functions.

Valeric and Valerko really want to watch the next Euro Championship game, so they asked you to help them.

Input

The first line contains integer n (1 ≤ n ≤ 105) — the number of functions. Each of the following n lines contains two space-separated integer numbers ki, bi ( - 109 ≤ ki, bi ≤ 109) that determine the i-th function.

Output

Print a single number — the number of angles that do not equal 180 degrees in the graph of the polyline that equals the sum of the given functions.

Input

1
1 0

Output

Input

Output

Input

Output

题目题意:问我们s(x)函数上有多少个不是180°的倾角。

题目分析:我们假象y(x)不是那样定义的，它就是简单的一次函数，那么n个一次函数相加肯定是一次函数，都是180°的倾角（直的），那么问题就出现在y(x)<0时，y(x)=0,也就是它本身应该加上一个小于0的数结果加上0了，所以就会有弯曲。

也就是记录与x轴有多少不同的交点

代码如下:

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<set>
#define ll long long
using namespace std;

set<long double> M;
int main()
{
    M.clear();
    int n;
    scanf("%d",&n);
    for (int i=1;i<=n;i++) {
        ll k,b;
        scanf("%lld%lld",&k,&b);
        if (k==0) continue;
        else {
            long double ans=-(long double)b/k;
            if (M.count(ans)==0) M.insert(ans);
        }
    }
    printf("%d\n",M.size());
    return 0;
}