Fruit Ninja(取随机数)

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

时间限制：C/C++ 5秒，其他语言10秒
空间限制：C/C++ 262144K，其他语言524288K
64bit IO Format: %lld

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 ≤ 10

⁴

) 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 x

_i

 and y

_i

 (-10

⁹

 ≤ x

_i

,y

_i

 ≤ 10

⁹

), 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".

输入例子:

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

输出例子:

Yes
No

-->

示例1

输入

复制

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

复制

Yes
No
思路：暴力必定超时，所以以取随机数的方式确定两个端点，然后从1到n枚举，看有多少个点在这条直线上，将此过程重复120次即可！
AC代码：

#include <bits/stdc++.h>
using namespace std;
int t,n,m,sum,x,y;
double k;
bool flag;
struct record
{
    int x,y;
};
record stu[];
int main()
{
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d %lf",&n,&k);
        for(int i=; i<=n-; i++)
        {
            scanf("%d %d",&stu[i].x,&stu[i].y);
        }
        if(n<=)
        {
            printf("Yes\n");
            continue;
        }
        m=;
        flag=false;
        while(m--)
        {
            y=rand()%n;
            x=rand()%n;
            if(x==y) continue;
            sum=;
            for(int i=; i<=n-; i++)
            {
                if(i==y || i==x)
                {
                    continue;
                }
                if((stu[i].y-stu[y].y)*(stu[i].x-stu[x].x)==(stu[i].y-stu[x].y)*(stu[i].x-stu[y].x))
                {
                    sum++;
                }
            }
            if((double)(sum)/n>=k)
            {
                flag=true;
                break;
            }
        }
        if(flag==true) printf("Yes\n");
        else printf("No\n");
    }
    return ;
}