目录
CSU 1007 矩形着色
题目：
Description
Danni想为屏幕上的一个矩形着色，但是她想到了一个问题。当点击鼠标以后电脑是如何判断填充的区域呢？现在给你一个平面直角坐标系，其中有一个矩形和一个点，矩形的四条边均是平行于x轴或y轴的。请你判断这个点相对于矩形的位置，即在矩形内，在矩形上，还是在矩形外?
Input
第一行只有一个整数T，(T < 150)，代表共有T种情况。
接下对于每种情况，均有两行数据：
第一行有两个整数Px Py，以空格分隔，代表点的坐标(Px,Py).
第二行有四个整数Ax Ay Bx By，以空格分隔，代表矩形左下角的坐标(Ax,Ay)和右上角的坐标(Bx,By).
所有的坐标均为区间[0,100]内的整数,且Ax<Bx，Ay<By
Output
对于每种情况仅输出一行：
1. 如果点在矩形外部，请输出”Outside”
2. 如果点正好在矩形的边上，请输出”On”
3. 如果点在矩形内部，请输出”Inside”
所有输出都不包含引号。
Sample Input
3
38 730 7 52 66
55 1
9 13 54 84
74 67
73 66 76 68
Sample Output
On
Outside
Inside
代码：
#include<iostream>
using namespace std;
int main()
{
int t;
cin >> t;
while (t--)
{
int px, py,ax, ay, bx, by;
cin >> px >> py >> ax >> ay >> bx >> by;
if (px<ax || px>bx || py<ay || py>by)cout << "Outside";
else if (px == ax || px == bx || py == ay || py == by)cout << "On";
else cout << "Inside";
cout << endl;
}
return 0;
}CSU 1011: Counting Pixels
题目：
Description
Did you know that if you draw a circle that fills the screen on your 1080p high definition display, almost a million pixels are lit? That's a lot of pixels! But do you know exactly how many pixels are lit? Let's find out!
Assume that our display is set on a Cartesian grid where every pixel is a perfect unit square. For example, one pixel occupies the area of a square with corners (0,0) and (1,1). A circle can be drawn by specifying its center in grid coordinates and its radius. On our display, a pixel is lit if any part of it is covered by the circle being drawn; pixels whose edge or corner are just touched by the circle, however, are not lit.
Your job is to compute the exact number of pixels that are lit when a circle with a given position and radius is drawn.
Input
The input consists of several test cases, each on a separate line. Each test case consists of three integers, x,y, and r(1≤x,y,r≤1,000,000), specifying respectively the center (x,y) and radius of the circle drawn. Input is followed by a single line with x = y = r = 0, which should not be processed.
Output
For each test case, output on a single line the number of pixels that are lit when the specified circle is drawn. Assume that the entire circle will fit within the area of the display.
Sample Input
1 1 1
5 2 5
0 0 0
Sample Output
4
88
思路：
x和y是没用的，直接暴力求解加起来就行
代码：
#include<iostream>
#include<math.h>
using namespace std;
int main()
{
long long r, ans;
while (cin >> r >> r >> r)
{
if (r == 0)break;
ans = 0;
for (long long i = 0; i < r; i++)
ans += int(sqrt(r*r - i * i) - 0.000000001) + 1;
cout << ans * 4 << endl;
}
return 0;
}CSU 1159: 中南才女
目：
Description
话说中南有位才女，叫做珊，学校里面很多同学都很喜欢她，想追她做女朋友。hkx也不例外。可是不知道为什么，向她表白的同学都一一落马，一向自卑的hkx一直也不敢表白，hkx得知珊有一天晚上8点钟以后的某段时间会去学校的某个地方弹吉他，可是hkx晚上9点以后才有时间去，虽然晚了一点，但也有机会，所以hkx就去了。
假设珊晚上到达的时刻是等可能的分布在8点到8+L点，hkx到达的时刻是等可能的分布在9点到9+L点。珊到达之后会在那里弹吉他t小时，之后就会离开。hkx到达之后如果没见到珊的话，他就会等上t小时，如果这t小时内珊还不出现，他就会离开。那么hkx能够见到珊的概率是多少呢？
Input
多组数据，每组数据包括两个实数，L和t，一组数据占一行
1<=L<=10
0<=t<=10
Output
输出hkx能够见到珊的概率，四舍五入到小数点后6位
Sample Input
2.0 0.0
2.0 3.0
Sample Output
0.000000
1.000000
这个题目其实好简单，就是在 1<y<1+L，0<x<L 这个正方形中，求满足 |y-x|<t 的部分所占的比例。
无非就是把直线y=x的下面和上面的部分分开求。
下面的部分是
sum = (l - 1)*(l - 1) / 2;
if (t < l - 1)sum -= (l - 1 - t)*(l - 1 - t) / 2;
上面的部分是
if (t < 1)sum += t*t / 2 + t*(l - 1);
else if(t<l+1)sum += l - 0.5 + l*l / 2 - (l - t + 1)*(l - t + 1) / 2;
else sum += l - 0.5 + l*l / 2;
完整代码：
#include<iostream>
#include<iomanip>
using namespace std;
int main()
{
double t, l, sum;
while(cin >> l >> t)
{
sum = (l - 1)*(l - 1) / 2;
if (t < l - 1)sum -= (l - 1 - t)*(l - 1 - t) / 2;
if (t < 1)sum += t*t / 2 + t*(l - 1);
else if(t<l+1)sum += l - 0.5 + l*l / 2 - (l - t + 1)*(l - t + 1) / 2;
else sum += l - 0.5 + l*l / 2;
cout << fixed << setprecision(6) << sum / l / l << endl;
}
return 0;
}CSU 1201: Triangle
题目：
Description
已知三角形各边边长的取值范围，你能求出三角形最大的面积吗？
Input
输入数据的第一行包含一个整数T (1 <= T <= 200)，表示接下来一共有T组测试数据。
每组数据占三行，每行均包含两个整数x, y (1 <= x <= y <= 1000)，表示该边边长可以为[x, y]范围内的任意实数。
Output
用一行输出三角形的最大的面积，结果保留3位小数。如果三条边不能构成三角形，则用一行输出“No Triangle”（不包括引号）。
Sample Input
2
1 2
1 2
4 5
1 3
2 3
3 3
Sample Output
No Triangle
3.897
代码：
#include<iostream>
#include<stdio.h>
#include<math.h>
using namespace std;
struct node
{
int x, y;
}nod[3];
double g(double a, double b, double c)
{
double p = (a + b + c) / 2;
return sqrt(p*(p - a)*(p - b)*(p - c));
}
double f(double a, int b, int x, int y)
{
if (x >= a + b) return -1;
double s = sqrt(a*a + b*b);
if (x > s)return g(a, b, x);
if (y < s)return g(a, b, y);
return a*b / 2;
}
int main()
{
int t;
scanf("%d", &t);
while (t--)
{
scanf("%d%d%d%d%d%d", &nod[0].x, &nod[0].y, &nod[1].x, &nod[1].y, &nod[2].x, &nod[2].y);
int i1 = 0, i2 = 1, i3 = 2;
if (nod[i1].y > nod[i2].y)i1 ^= i2 ^= i1 ^= i2;
if (nod[i1].y > nod[i3].y)i1 ^= i3 ^= i1 ^= i3;
if (nod[i2].y > nod[i3].y)i2 ^= i3 ^= i2 ^= i3;
double r = f(nod[i1].y, nod[i2].y, nod[i3].x, nod[i3].y);
if (r>-1)printf("%.3f\n", r);
else printf("No Triangle\n");
}
return 0;
}CSU 1204 Rectangles
题目：
Description
如果限定矩形的边长必须为整数，且周长为定值L，那么面积在[A, B]范围内不同的矩形一共有多少个呢？
在这个问题中，当且仅当两个矩形面积不同时，视作是两个不同的矩形。
Input
输入数据的第一行包含一个整数T (1 <= T <= 10000)，表示接下来一共有T组测试数据。
对于每组测试数据，包含三个整数L (1 <= L <= 2,000,000,000)、A、B (1 <= A <= B <= 250,000,000,000,000,000)，含义同上。
Output
对于每组测试数据，用一行输出一个整数，代表上述问题的答案。
Sample Input
3
11 1 6
12 5 10
12 8 10
Sample Output
0
3
2
这个题目，很明显是必须直接求出表达式，因为给的数太大了。
代码：
#include<iostream>
#include<math.h>
using namespace std;
int main()
{
int T;
cin >> T;
long long a, b, l, max, min;
while (T--)
{
cin >> l >> a >> b;
if (l % 2)cout << 0 << endl;
else
{
l = l / 2;
max = l / 2;
if (max*(l - max) > b)
{
max = (long long)(l*1.0 / 2 - sqrt(l*l*1.0 / 4 - b));
if (max*(l - max) > b)max -= 1;
else if ((max + 1)*(l - max - 1) <= b && max < l / 2)max += 1;
}
min = (l + 1) / 2;
if (min*(l - min) == a)
{
cout << 1 << endl;
continue;
}
if (min*(l - min) < a)
{
cout << 0 << endl;
continue;
}
else
{
min = (long long)(l*1.0 / 2 - sqrt(l*l*1.0 / 4 - a)) + 1;
if ((min - 1)*(l - min + 1) >= a)min -= 1;
else if (min*(l - min)<a)min += 1;
}
if (max && min <= max)cout << max - min + 1 << endl;
else cout << 0 << endl;
}
}
return 0;
}
CSU 1403: 三角形面积
题目：
Description
给出三维空间上的三个点，计算以这三个点为顶点的三角形的面积。
Input
输入的第一行包含一个整数T (T > 0)，表示一共有T组测试数据。
每组数据占三行，每行各描述了一个点的坐标。点的坐标会以x y z的形式给出，表示这个点的坐标为(x, y, z)，其中x, y, z均为[-100, 100]范围内的浮点数。
Output
对于每组测试数据，输出三角形的面积（和标准答案的误差不超过10-6即可）。
Sample Input
2
0 1 0
1 0 0
0 0 0
0 0.0 1.00
1.0 0.00 0
0.00 1.0 0
Sample Output
0.50
0.8660254
代码：
#include<iostream>
#include<math.h>
#include<iomanip>
using namespace std;
int main()
{
int t;
double xa, ya, za, xb, yb, zb, xc, yc, zc;
double l1, l2, l3, p;
cin >> t;
while (t--)
{
cin >> xa >> ya >> za >> xb >> yb >> zb >> xc >> yc >> zc;
l1 = sqrt((xa - xb)*(xa - xb) + (ya - yb)*(ya - yb) + (za - zb)*(za - zb));
l2 = sqrt((xa - xc)*(xa - xc) + (ya - yc)*(ya - yc) + (za - zc)*(za - zc));
l3 = sqrt((xb - xc)*(xb - xc) + (yb - yc)*(yb - yc) + (zb - zc)*(zb - zc));
p = (l1 + l2 + l3) / 2;
cout << setprecision(16) << sqrt(p*(p - l1)*(p - l2)*(p - l3)) << endl;
}
return 0;
}CSU 1407: 最短距离
题目：
Description
两个点 A , B 均在做匀速直线运动。给出 t = 0时刻 A , B 的坐标，以及 A , B 的速度，计算 t ≥ 0时两个点的距离的最小值。
Input
输入的第一行包含一个整数 T (1 ≤ T ≤ 200 )，表示一共有 T 组测试数据。
对于每组测试数据，第一行包含4个整数 x A , y A , v Ax , v Ay (-10 3 ≤ x A , y A , v Ax , v Ay ≤ 10 3 )，表示 t = 0时刻 A 的坐标为( x A , y A )， A 的速度在 x 轴方向上的分量为 v Ax ，在 y 轴上的分量为 v Ay 。第二行包含四个整数 x B , y B , v Bx , v By (-10 3 ≤ x B , y B , v Bx , v By ≤ 10 3 )，以相同的方式给出了 B 的各项属性。
Output
对于每组测试数据，输出 t ≥ 0时两个点距离的最小值，保留8位小数。
Sample Input
6
0 0 0 0
0 1 0 1
0 0 -1 1
0 0 1 -1
0 1 1 0
2 0 0 1
0 1 1 0
2 0 1 0
0 0 -1 1
1 1 1 -1
997 997 -1000 -1000
-1000 -1000 1000 1000
Sample Output
1.00000000
0.00000000
0.70710678
2.23606798
1.41421356
0.00000000
代码：
#include<iostream>
#include<math.h>
#include<iomanip>
using namespace std;
int main()
{
int t;
double xa, ya, vxa, vya;
double xb, yb, vxb, vyb;
double a, b, c;//att+2bt+c
cin >> t;
while (t--)
{
cin >> xa >> ya >> vxa >> vya;
cin >> xb >> yb >> vxb >> vyb;
a = (vxa - vxb)*(vxa - vxb) + (vya - vyb)*(vya - vyb);
b = (vxa - vxb)*(xa - xb) + (vya - vyb)*(ya - yb);
c = (xa - xb)*(xa - xb) + (ya - yb)*(ya - yb);
cout << setprecision(10) << sqrt((a*b >= 0) ? c : c - b * b / a) << endl;
}
return 0;
}CSU 1591: 三角形
题目：
Description
star有n根火柴棒。由于他的火柴棒不是制式的，所以长度可能会不一样。现在star想选三根火柴棒搭一个三角形送给女朋友，并且希望三角形的面积尽量大。现在希望你编程告诉star可能搭成的最大三角形的面积是多少。
Input
第一行只有一个整数T(1<=T<=20)，表示数据组数。
下面的T行每一行有一个整数n(3<=n<=100)，表示火柴棒的数量，后面跟着n个整数Li(1<=Li<=100)，表示n跟火柴棒的长度。
Output
对于每一组数据输出一个浮点数S，表示最大面积。
结果保留三位小数(四舍五入)。
如果无法搭成三角形，输出”no”。
Sample Input
3
3 3 4 5
4 3 4 5 5
3 1 1 2
Sample Output
6.000
9.165
no
代码：
#include<iostream>
#include<math.h>
#include<stdio.h>
using namespace std;
double f(int a, int b, int c)
{
if (a + b <= c || a + c <= b || b + c <= a)return 0;
double p = (a + b + c)*0.5;
return sqrt(p*(p - a)*(p - b)*(p - c));
}
int main()
{
int t, n, len[100];
cin >> t;
while (t--)
{
cin >> n;
for (int i = 0; i < n; i++)cin >> len[i];
double max = 0;
for (int i = 0; i < n; i++)for (int j = i + 1; j < n; j++)for (int k = j + 1; k < n; k++)
if (max < f(len[i], len[j], len[k]))max = f(len[i], len[j], len[k]);
if (max)printf("%.3f\n", max);
else cout << "no\n";
}
return 0;
}CSU 1896: Symmetry（对称点集）
题目：
Description
We call a figure made of points is left-right symmetric as it is possible to fold the sheet of paper along a vertical line and to cut the figure into two identical halves.For example, if a figure exists five points, which respectively are (-2,5),(0,0),(2,3),(4,0),(6,5). Then we can find a vertical line x = 2 to satisfy this condition. But in another figure which are (0,0),(2,0),(2,2),(4,2), we can not find a vertical line to make this figure left-right symmetric.Write a program that determines whether a figure, drawn with dots, is left-right symmetric or not. The dots are all distinct.
Input
The input consists of T test cases. The number of test cases T is given in the first line of the input file. The first line of each test case contains an integer N, where N (1 <=N <= 1, 000) is the number of dots in a figure. Each of the following N lines contains the x-coordinate and y-coordinate of a dot. Both x-coordinates and y-coordinates are integers between −10, 000 and 10, 000, both inclusive.
Output
Print exactly one line for each test case. The line should contain 'YES' if the figure is left-right symmetric,and 'NO', otherwise.
Sample Input
3
5
-2 5
0 0
6 5
4 0
2 3
4
2 3
0 4
4 0
0 0
4
5 14
6 10
5 10
6 14
Sample Output
YES
NO
YES
这个题目是说，给一个整点集，判断是不是左右对称的。
要分2步进行，第一步找到唯一可能的对称轴的位置，第二步是判断点集是不是关于这条直线对称的。
第一步我是找最中间的若干个点，第二步我是把所有不在对称轴上面的点映射到一个整数，使得对称的点映射到一样的整数，不对称的点几乎不会映射到一样的整数，最后求出所有整数的异或，是0就对称，不是0就不对称。
我在比赛时提交的代码：
#include<iostream>
#include<algorithm>
using namespace std;
int kx1,kx2;
struct node
{
int x,y;
};
bool cmp(node n1,node n2)
{
return n1.x<n2.x;
}
int f(node nodd,int kx)
{
if(nodd.x*2<kx)return (kx-nodd.x*2)%123*nodd.y;
return (nodd.x*2-kx)%123*nodd.y;
}
int main()
{
int T,N,ans1,ans2;
node nod[1000];
cin>>T;
while(T--)
{
cin>>N;
for(int i=0;i<N;i++)cin>>nod[i].x>>nod[i].y;
sort(nod,nod+N,cmp);
kx1=nod[N/2].x*2;
kx2=nod[N/2].x+nod[(N-1)/2].x;
ans1=ans2=0;
for(int i=0;i<N;i++)
{
ans2=ans2^f(nod[i],kx2);
if(nod[i].x==kx1)continue;
ans1=ans1^f(nod[i],kx1);
}
if(ans1==0||ans2==0)cout<<"YES\n";
else cout<<"NO\n";
}
return 0;
}实际上求对称轴还有更好的方法，不用最中间的点，用最左边和最右边的点。
第二步也不需要映射到整数，直接把点集存2份，一份从左往右排序，一份从右往左排序，然后一一比较是否对称即可。
新代码：
#include<iostream>
#include<algorithm>
using namespace std;
int T, N, kx;
struct node
{
int x, y;
};
node nod1[1000], nod2[1000];
bool cmp1(node n1, node n2)
{
if (n1.x == n2.x)return n1.y < n2.y;
return n1.x<n2.x;
}
bool cmp2(node n1, node n2)
{
if (n1.x == n2.x)return n1.y < n2.y;
return n1.x>n2.x;
}
bool ok()
{
for (int i = 0; i < N; i++)
{
if (nod1[i].y != nod2[i].y)return false;
if (nod1[i].x + nod2[i].x != kx)return false;
}
return true;
}
int main()
{
int left, right;
cin >> T;
while (T--)
{
cin >> N;
left = 10000, right = -10000;
for (int i = 0; i < N; i++)
{
cin >> nod1[i].x >> nod1[i].y;
nod2[i].x = nod1[i].x, nod2[i].y = nod1[i].y;
if (left>nod1[i].x)left = nod1[i].x;
if (right < nod1[i].x)right = nod1[i].x;
}
kx = left + right;
sort(nod1, nod1 + N, cmp1);
sort(nod2, nod2 + N, cmp2);
if (ok())cout << "YES\n";
else cout << "NO\n";
}
return 0;
}POJ 202 Squares
题目：
Description
A square is a 4-sided polygon whose sides have equal length and adjacent sides form 90-degree angles. It is also a polygon such that rotating about its centre by 90 degrees gives the same polygon. It is not the only polygon with the latter property, however, as a regular octagon also has this property.
So we all know what a square looks like, but can we find all possible squares that can be formed from a set of stars in a night sky? To make the problem easier, we will assume that the night sky is a 2-dimensional plane, and each star is specified by its x and y coordinates.
Input
The input consists of a number of test cases. Each test case starts with the integer n (1 <= n <= 1000) indicating the number of points to follow. Each of the next n lines specify the x and y coordinates (two integers) of each point. You may assume that the points are distinct and the magnitudes of the coordinates are less than 20000. The input is terminated when n = 0.
Output
For each test case, print on a line the number of squares one can form from the given stars.
Sample Input
4
1 0
0 1
1 1
0 0
9
0 0
1 0
2 0
0 2
1 2
2 2
0 1
1 1
2 1
4
-2 5
3 7
0 0
5 2
0
Sample Output
1
6
1
我感觉自己的做法挺好的啊啊，然而超时了
超时代码：
#include<iostream>
#include<algorithm>
#include<cmath>
#include<string.h>
using namespace std;
struct node
{
long long sum, dx, dy;
};
long long point[1000];
long long h = 40001;
node nod[1000000];
bool cmp(node a, node b)
{
return a.sum < b.sum;
}
long long getx(long long m)
{
if (m >= 0)return m / h;
else return (m + 1) / h - 1;
}
long long f(long long a, long long b) //屏蔽掉对称性
{
if (a > b)return a*h + b;
return a + b*h;
}
int main()
{
ios_base::sync_with_stdio(false);
int n, x, y;
while (cin >> n)
{
if (n == 0)break;
for (int i = 0; i < n; i++)
{
cin >> x >> y;
point[i] = x*h + y + h / 2; //y + h / 2是0到4万， x*h + y + h / 2是大约-8亿到8亿
}
for (int i = 0; i < n*n; i++)
{
nod[i].sum = h *h * 2000000000;
nod[i].dx = 0;
nod[i].dy = 0;
}
for (int i = 0; i < n; i++)for (int j = i + 1; j < n; j++)
{
nod[i*n + j].dx = getx(point[i]) - getx(point[j]);
nod[i*n + j].dy = point[i] - getx(point[i])*h - (point[j] - getx(point[j])*h);
nod[i*n + j].sum = (point[i] + point[j])*h*h + f(abs(int(nod[i*n + j].dx)), abs(int(nod[i*n + j].dy)));//屏蔽掉正负
}
sort(nod, nod + n*n, cmp);
int j, num = 0;
for (int i = 1; i <= n*n; i++)
{
if (nod[i].sum == h *h * 2000000000)break;
if (nod[i].sum == nod[i - 1].sum) //满足这个的最多只有8种情况
{
j = i - 1;
while (i <= n*n && nod[i].sum == nod[i - 1].sum)
{
for (int k = j; k < i; k++)
{
if (nod[k].dx == nod[i].dy && nod[k].dy == -nod[i].dx)num++;
if (nod[k].dy == nod[i].dx && nod[k].dx == -nod[i].dy)num++;
}
i++;
}
}
}
cout << num << endl;
}
return 0;
}POJ 2002 Squares
题目：
A square is a 4-sided polygon whose sides have equal length and adjacent sides form 90-degree angles. It is also a polygon such that rotating about its centre by 90 degrees gives the same polygon. It is not the only polygon with the latter property, however, as a regular octagon also has this property.
So we all know what a square looks like, but can we find all possible squares that can be formed from a set of stars in a night sky? To make the problem easier, we will assume that the night sky is a 2-dimensional plane, and each star is specified by its x and y coordinates.
Input
The input consists of a number of test cases. Each test case starts with the integer n (1 <= n <= 1000) indicating the number of points to follow. Each of the next n lines specify the x and y coordinates (two integers) of each point. You may assume that the points are distinct and the magnitudes of the coordinates are less than 20000. The input is terminated when n = 0.
Output
For each test case, print on a line the number of squares one can form from the given stars.
Sample Input
4
1 0
0 1
1 1
0 0
9
0 0
1 0
2 0
0 2
1 2
2 2
0 1
1 1
2 1
4
-2 5
3 7
0 0
5 2
0
Sample Output
1
6
1
题意：给出n个点，求出正方形数量
思路：枚举每2个点组成的线段，判断是不是正方形的对角线
代码：
#include<iostream>
#include<map>
using namespace std;
int N = 50000;
struct node
{
int dx;
int dy;
}nod[1005];
map<int,int>m;
int main()
{
ios_base::sync_with_stdio(false);
int n, x, y;
while (cin >> n)
{
m.clear();
if (n == 0)break;
for (int i = 0; i < n; i++)
{
cin >> nod[i].dx >> nod[i].dy;
m.insert(pair<int,int>(nod[i].dx*N+nod[i].dy,1));
}
int sum = 0;
for (int i = 0; i < n; i++)for (int j = i+1; j < n; j++)
{
x = nod[i].dx + nod[j].dx + nod[i].dy - nod[j].dy;
y = nod[i].dy + nod[j].dy + nod[j].dx - nod[i].dx;
if (x % 2 || y % 2)continue;
x /= 2, y /= 2;
if (x > 20000 || x < -20000 || y>20000 || y < -20000)continue;
if (m.find(x*N + y) == m.end())continue;
x = nod[i].dx + nod[j].dx + nod[j].dy - nod[i].dy;
y = nod[i].dy + nod[j].dy + nod[i].dx - nod[j].dx;
if (x % 2 || y % 2)continue;
x /= 2, y /= 2;
if (x > 20000 || x < -20000 || y>20000 || y < -20000)continue;
if (m.find(x*N + y) == m.end())continue;
sum++;
}
cout << sum/2 << endl;
}
return 0;
}