P1355 神秘大三角

合集 - 模拟(54)

1.G. Vlad and Trouble at MIT2024-02-202.P2899 [USACO08JAN] Cell Phone Network G2024-02-203.Watering an Array2024-02-234.P2055 [ZJOI2009] 假期的宿舍2024-03-085.T432579 大西洋船王2024-03-086.T432580 星云 hard ver.2024-03-087.P7072 [CSP-J2020] 直播获奖2024-03-098.E - Insert or Erase2024-03-109.D - String Bags2024-03-1010.P2671 [NOIP2015 普及组] 求和2024-03-1011.P1884 [USACO12FEB] Overplanting S2024-03-1012.F. Microcycle2024-03-1213.E. Increasing Subsequences__22024-03-1314.G. Path Prefixes2024-03-1415.F. Rudolf and Imbalance2024-03-14

16.P1355 神秘大三角2024-03-20

17.[ARC172A] Chocolate2024-03-2518.F. 0, 1, 2, Tree!2024-03-2919.E. Nearly Shortest Repeating Substring2024-03-2920.P1354 房间最短路问题2024-03-2921.P3258 [JLOI2014] 松鼠的新家2024-04-0222.P3052 [USACO12MAR] Cows in a Skyscraper G2024-04-0523.P1337 [JSOI2004] 平衡点 / 吊打XXX2024-04-0724.P3745 [六省联考 2017] 期末考试2024-04-0825.D. Inaccurate Subsequence Search2024-04-0926.E. Long Inversions2024-04-0927.C. Deep Down Below2024-04-0928.C. Rings2024-04-0929.F. Unfair Game2024-04-1030.C. Inversion Graph2024-04-1031.D. Cross Coloring2024-04-1232.D2. Set To Max (Hard Version)2024-04-1233.C. Array Game2024-04-1234.F. Alex's whims2024-04-1235.Queue Sort2024-04-1236.P10252 线性变换2024-04-1637.P10288 [GESP样题 八级] 区间2024-04-1638.P2161 [SHOI2009] 会场预约2024-04-1739.L2-001 紧急救援2024-04-1840.L2-002 链表去重2024-04-1841.L2-013 红色警报2024-04-1842.L2-014 列车调度2024-04-1943.L2-022 重排链表2024-04-1944.L2-023 图着色问题2024-04-1945.L2-039 清点代码库2024-04-1946.L2-034 口罩发放2024-04-1947.L3-037 夺宝大赛2024-04-2348.仓库规划2024-04-2349.因子化简2024-04-2350.C. Torn Lucky Ticket2024-04-2751.俄罗斯方块2024-05-1452.F. Cutting Game2024-05-2153.D. Prefix Permutation Sums2024-06-1754.P3304 [SDOI2013] 直径2024-07-15

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原题链接

题解

叉积的运用,scanf控制输入格式

code

#include<bits/stdc++.h>
using namespace std;
struct node
{
    int x,y;
}a[100005];
int x[200005],y[200005];
int main()
{
    for(int i=0;i<3;i++)
    {
        scanf("(%d,%d)\n",&a[i].x,&a[i].y);
    }

    int x1,y1;
    scanf("(%d,%d)",&x1,&y1);

    for(int i=0;i<3;i++)
    {
        x[i]=a[i].x-x1,y[i]=a[i].y-y1;
    }

    int zero=0,fu=0,zheng=0;
    for(int i=0;i<3;i++)
    {
        if(x[i]*y[(i+1)%3]-x[(i+1)%3]*y[i]==0)
        {
            zero++;
        }
        else if(x[i]*y[(i+1)%3]-x[(i+1)%3]*y[i]<0)
        {
            fu++;
        }
        else zheng++;
    }

    if(fu&&zheng) cout<<2;
    else if(zero==1) cout<<3;
    else if(zero==2) cout<<4;
    else cout<<1;
    return 0;
}