zoukankan      html  css  js  c++  java
  • B. Tell Your World(几何数学 + 思维)

    B. Tell Your World
    time limit per test
    1 second
    memory limit per test
    256 megabytes
    input
    standard input
    output
    standard output

    Connect the countless points with lines, till we reach the faraway yonder.

    There are n points on a coordinate plane, the i-th of which being (i, yi).

    Determine whether it's possible to draw two parallel and non-overlapping lines, such that every point in the set lies on exactly one of them, and each of them passes through at least one point in the set.

    Input

    The first line of input contains a positive integer n (3 ≤ n ≤ 1 000) — the number of points.

    The second line contains n space-separated integers y1, y2, ..., yn ( - 109 ≤ yi ≤ 109) — the vertical coordinates of each point.

    Output

    Output "Yes" (without quotes) if it's possible to fulfill the requirements, and "No" otherwise.

    You can print each letter in any case (upper or lower).

    Examples
    input
    Copy
    5
    7 5 8 6 9
    output
    Copy
    Yes
    input
    Copy
    5
    -1 -2 0 0 -5
    output
    Copy
    No
    input
    Copy
    5
    5 4 3 2 1
    output
    Copy
    No
    input
    Copy
    5
    1000000000 0 0 0 0
    output
    Copy
    Yes
    Note

    In the first example, there are five points: (1, 7), (2, 5), (3, 8), (4, 6) and (5, 9). It's possible to draw a line that passes through points 1, 3, 5, and another one that passes through points 2, 4 and is parallel to the first one.

    In the second example, while it's possible to draw two lines that cover all points, they cannot be made parallel.

    In the third example, it's impossible to satisfy both requirements at the same time.

    算法:几何数学 + 思维

    #include <iostream>
    #include <cstdio>
    #include <algorithm>
    
    using namespace std;
    
    typedef long long ll;
    
    #define INF 0x3f3f3f3f
    const int maxn = 1e5+7;
    
    ll a[maxn];
    int n;
    
    int solve(double k) {
        int pos = -1;
        for(int i = 2; i <= n; i++) {
            if(a[i] - a[1] == (i - 1) * k ) {
                continue;
            }
            if(pos == -1) {
                pos = i;        //确定一个新的基点
            } else if(a[i] - a[pos] != (i - pos) * k){
                return 0;
            }
        }
        return pos != -1;   //判断是否是所有的点都在一条直线上
    }
    
    int main() {
        while(~scanf("%d", &n)) {
            for(int i = 1; i <= n; i++) {
                cin >> a[i];
            }
            //以三点来确定三条直线,有以下三种情况
            double k1 = a[2] - a[1];        
            double k2 = 1.0 * (a[3] - a[1]) / 2;
            double k3 = a[3] - a[2];
            if(solve(k1) || solve(k2) || solve(k3)) {
                printf("Yes
    ");
            } else {
                printf("No
    ");
            }
        }
        return 0;
    }
  • 相关阅读:
    L3-007. 天梯地图
    智能算法-遗传算法
    L2-012. 关于堆的判断
    L2-010. 排座位
    计算几何初步-三点顺序
    L3-001. 凑零钱
    靠二进制画几何[图论]
    【排序】
    欧拉路与欧拉回路
    Test on 2016/09/26
  • 原文地址:https://www.cnblogs.com/buhuiflydepig/p/11312083.html
Copyright © 2011-2022 走看看