2018 ACM 国际大学生程序设计竞赛上海大都会赛重现赛A: Fruit Ninja(随机数)

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

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.c++

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1git

输入

复制ide

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ui

Yes
No

 

没啥可说的太菜想不到随机数,想到很简单。this

 

 

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
#define IO ios::sync_with_stdio(false),cin.tie(0)
#define FIN freopen("D://code//in.txt", "r", stdin)
#define ppr(i,x,n) for(int i = x;i <= n;i++)
#define rpp(i,n,x) for(int i = n;i >= x;i--)
const double eps = 1e-8;
const int mod = 1e9 + 7;
const int maxn = 1e7 + 10;
const double pi = acos(-1);
const int inf = 0x3f3f3f3f;
const ll INF = 0x3f3f3f3f3f3f3f3f;
inline ll qow(ll a,ll b){ll r=1,t=a; while(b){if(b&1)r=(r*t)%mod;b>>=1;t=(t*t)%mod;}return r;}
inline int read() {//读入挂
    int ret = 0, c, f = 1;
    for(c = getchar(); !(isdigit(c) || c == '-'); c = getchar());
    if(c == '-') f = -1, c = getchar();
    for(; isdigit(c); c = getchar()) ret = ret * 10 + c - '0';
    if(f < 0) ret = -ret;
    return ret;
}
int x[maxn],y[maxn];
int main()
{
	int t;
	scanf("%d",&t);
	srand(time(0));
	while(t--)
	{
		int n;bool flag = false;
		double d;
		scanf("%d%lf",&n,&d);
		ppr(i,0,n-1)scanf("%d%d",x+i,y+i);
		ppr(i,1,999)
		{
			int countt = 0;int a = rand()%n;int b = rand()%n;
			while(a == b) b = rand()%n;
			ppr(j,0,n-1)
			{
				if((y[j]-y[a])*(x[b]-x[j]) == (y[b]-y[j])*(x[j]-x[a]))
				countt++;
			}
			if(countt >= n*d)
			{
				flag = true;
				break;
			}
		}
		if(flag)
		printf("Yes\n");
		else
		printf("No\n");
	}
	return 0;
}