Fruit Ninja

链接：https://www.nowcoder.com/acm/contest/163/A
来源：牛客网
 

题目描述

Fruit Ninja is a juicy action game enjoyed by millions of players around the world, with squishy,
splat and satisfying fruit carnage! Become the ultimate bringer of sweet, tasty destruction with every slash.
Fruit Ninja is a very popular game on cell phones where people can enjoy cutting the fruit by touching the screen.
In this problem, the screen is rectangular, and all the fruits can be considered as a point. A touch is a straight line cutting
thought the whole screen, all the fruits in the line will be cut.
A touch is EXCELLENT if ≥ x, (N is total number of fruits in the screen, M is the number of fruits that cut by the touch, x is a real number.)
Now you are given N fruits position in the screen, you want to know if exist a EXCELLENT touch.

输入描述:

The first line of the input is T(1≤ T ≤ 100), which stands for the number of test cases you need to solve.
The first line of each case contains an integer N (1 ≤ N ≤ 104) and a real number x (0 < x < 1), as mentioned above.
The real number will have only 1 digit after the decimal point.
The next N lines, each lines contains two integers xi and yi (-109 ≤ xi,yi ≤ 109), denotes the coordinates of a fruit.

输出描述:

For each test case, output "Yes" if there are at least one EXCELLENT touch. Otherwise, output "No".

 

示例1

输入

2
5 0.6
-1 -1
20 1
1 20
5 5
9 9
5 0.5
-1 -1
20 1
1 20
2 5
9 9

输出

Yes
No

题解：要求找出最多的在一条直线上的点是否大于等于总数乘以某个比例；

           随机从n个点里枚举两个点，看两个点是否能和其他的点斜率相同，当相同的个数超过比例数时就符合条件；

           随机的次数，我随机了1000个点，这样过的概率就很大了，应该不会wa；

#include<bits/stdc++.h>
using namespace std;
struct Point
{
    int x,y;
}p[10005];
int judge(int x,int y,int a,int b)
{
    return a*y-b*x;
}
int main()
{
    int n,a,b,t,ans;
    double mod;
    cin>>t;
    while(t--)
    {
        cin>>n>>mod;
        for(int i=0;i<n;i++)
        {
            cin>>p[i].x>>p[i].y;
        }
        int sum=1001;
        while(sum--)
        {
            ans=0;
            a=rand()%n;//随机生成从0~n-1，之间的数
            b=rand()%n;
            if(a==b)
            {
                continue;
            }
            for(int i=0;i<n;i++)
            {
                if(judge(p[i].x-p[a].x,p[i].y-p[a].y,p[b].x-p[a].x,p[b].y-p[a].y)==0)//斜率比较
                {
                    ans++;
                }
            }
            if(ans*100>=100*n*mod)//看总数是否符合
            {
                cout<<"Yes"<<endl;
                break;
            }
        }
        if(sum==-1)//当随机了1000次后还没有符合的就输出No
        {
            cout<<"No"<<endl;
        }
    }
}