Description

Stan has n sticks of various length. He throws them one at a time on the floor in a random way. After finishing throwing, Stan tries to find the top sticks, that is these sticks such that there is no stick on top of them. Stan has noticed that the last thrown stick is always on top but he wants to know all the sticks that are on top. Stan sticks are very, very thin such that their thickness can be neglected.                                                             
                                                                                                                                                                                      

Input

Output

Sample Input

5
1 1 4 2
2 3 3 1
1 -2.0 8 4
1 4 8 2
3 3 6 -2.0
3
0 0 1 1
1 0 2 1
2 0 3 1
0

Sample Output

Top sticks: 2, 4, 5.
Top sticks: 1, 2, 3.

Hint

/**//***********************************
暴力就行，从第一个开始判断
如果两条线段相交就把前面一条筛选掉
判断线段相交直接贴的吉大模板。。。
***********************************/
#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

const int maxn = 100000 + 5;
const double eps=1e-10;

struct point { double x, y; };

point p[maxn], b[maxn];
bool ans[maxn];

double min(double a, double b) { return a < b ? a : b; }

double max(double a, double b) { return a > b ? a : b; }

bool inter(point a, point b, point c, point d)
{
    if( min(a.x, b.x) > max(c.x, d.x) ||
        min(a.y, b.y) > max(c.y, d.y) ||
        min(c.x, d.x) > max(a.x, b.x) ||
        min(c.y, d.y) > max(a.y, b.y) )
    return 0;

    double h, i, j, k;

    h = (b.x - a.x) * (c.y - a.y) - (b.y - a.y) * (c.x - a.x);
    i = (b.x - a.x) * (d.y - a.y) - (b.y - a.y) * (d.x - a.x);
    j = (d.x - c.x) * (a.y - c.y) - (d.y - c.y) * (a.x - c.x);
    k = (d.x - c.x) * (b.y - c.y) - (d.y - c.y) * (b.x - c.x);

    return h * i <= eps && j * k <= eps;
}

int main()
{
    int n;
    int res[maxn];
    while( cin >> n, n )
    {
        memset( ans, 0, sizeof( ans ) );
        for( int i = 0; i < n; i++ )
        {
            cin >> p[i].x >> p[i].y >> b[i].x >> b[i].y;
        }

        for( int i = 0; i < n; i++ )
        {
            for( int j = i + 1; j < n; j++ )
            {
                if( inter(p[i], b[i], p[j], b[j] ) )
                {
                    ans[i] = 1;
                    break;              //不加break会超时。。。
                }
            }
        }
        int ct = 0;
        cout << "Top sticks: ";
        for( int i = 0; i < n; i++ )
            if( !ans[i] )  res[ct++] = i + 1;
        for( int i = 0; i < ct - 1; i++ )
            cout << res[i] << ", ";
        cout << res[ct-1] << "." << endl;
    }
    return 0;
}

Description

Input

Output

Sample Input

5
0 0 4 4 0 4 4 0
5 0 7 6 1 0 2 3
5 0 7 6 3 -6 4 -3
2 0 2 27 1 5 18 5
0 3 4 0 1 2 2 5

Sample Output

INTERSECTING LINES OUTPUT
POINT 2.00 2.00
NONE
LINE
POINT 2.00 5.00
POINT 1.07 2.20
END OF OUTPUT

/**//*************************************
计算几何基础题，判断直线相交及求交点
注意斜率不存在的情况
**************************************/
#include <iostream>
#include <cstdio>

int main()
{
    double x1, y1, x2, y2, x3, y3, x4, y4;
    int n;
    double k1, k2;
    double b1, b2;
    double i_x, i_y;
    scanf("%d", &n);
    std::cout << "INTERSECTING LINES OUTPUT" << std::endl;
    while( n-- )
    {
        scanf("%lf%lf%lf%lf%lf%lf%lf%lf", &x1, &y1, &x2, &y2, &x3, &y3, &x4, &y4);

        if( x1 != x2 && x3 != x4 )
        {
            k1 = ( y2 - y1 ) / ( x2 - x1 );
            k2 = ( y4 - y3 ) / ( x4 - x3 );
            b1 = y1 - k1 * x1;
            b2 = y3 - k2 * x3;
            if( k1 == k2 )
            {
                if( b1 == b2 )
                    printf("LINE\n");
                else
                    printf("NONE\n");
            }
            else
            {
                i_x = (b2 - b1) / (k1 - k2);
                i_y = k1 * i_x + b1;
                printf("POINT %.2lf %.2lf\n", i_x, i_y);
            }
        }
        else if( x1 == x2 && x3 == x4 )
        {
            if( x1 == x3 )
            std::cout << "LINE\n";
            else
            std::cout << "NONE\n";
        }
        else if( x1 == x2 && x3 != x4 )
        {
            k2 = ( y4 - y3 ) / ( x4 - x3 );
            b2 = y3 - k2 * x3;
            i_x = x1;
            i_y = k2 * x1 + b2;
            printf("POINT %.2lf %.2lf\n", i_x, i_y);
        }
        else
        {
            k1 = ( y2 - y1 ) / ( x2 - x1 );
            b1 = y1 - k1 * x1;
            i_x = x3;
            i_y = k1 * x3 + b1;
            printf("POINT %.2lf %.2lf\n", i_x, i_y);
        }
    }
    std::cout << "END OF OUTPUT\n";
    return 0;
}

Description

Input

Output

Sample Input

95.123 12
0.4321 20
5.1234 15
6.7592  9
98.999 10
1.0100 12

Sample Output

548815620517731830194541.899025343415715973535967221869852721
.00000005148554641076956121994511276767154838481760200726351203835429763013462401
43992025569.928573701266488041146654993318703707511666295476720493953024
29448126.764121021618164430206909037173276672
90429072743629540498.107596019456651774561044010001
1.126825030131969720661201

虽然效率低点儿，但代码非常简单，容易实现，真正比赛还是很好用的。。。
import java.io.*;
import java.util.*;
import java.math.*;

public class Main{
    public static void main( String args[] )
    {
        BigDecimal num;
        int n;
        String r;
        Scanner cin = new Scanner(System.in);
        
        while(cin.hasNextBigDecimal())
        {
            num = cin.nextBigDecimal();
            n = cin.nextInt();
            
            num = num.pow(n);
            r = num.stripTrailingZeros().toPlainString();//BigDecimal.toPlainString 避免输出时产生科学计数法形式
            if(r.startsWith("0."))
                r = r.substring(1);
            System.out.println(r);
        }
    }
}