SWITCH

題目描述如下：

There are N lights in a line. Given the states (on/off) of the lights, your task is to determine at least how many lights should be switched (from on to off, or from off to on), in order to make the lights on and off alternatively.
Input
One line for each testcase.
The integer N (1 <= N <= 10000) comes first and is followed by N integers representing the states of the lights ("1" for on and "0" for off).
Process to the end-of-file.
Output
For each testcase output a line consists of only the least times of switches.
Sample Input
3 1 1 1
3 1 0 1
Sample Output
1
0

分析：該題看似簡單但卻隱藏著陷阱，題目要求尋找的是最少的切換數，故從第二盞燈開始判斷處理得出的結論是不一定正確的。通過分析可以發現該題其實只存在兩種情況：奇數位置的燈開著或者偶數位置的燈開著。這樣可以直觀的處理該題：取奇數位置燈開著需要切換燈狀態數與偶數位置燈開著需切換燈狀態數的較小值。這樣的話需要掃描兩邊燈的狀態數組，開銷較大。進一步分析，設a為奇數位置的燈開著需要切換的燈數，b為偶數位置燈開著需要切換的燈數。其實a+b=n。這樣本題就只需要掃描一遍數組，且進一步優化后存儲燈狀態的數組也可以省了。具體代碼如下：

 1#include <stdio.h>
 2#include <stdlib.h>
 3
 4int main(void)
 5{
 6    int n;
 7    int prev;
 8    int tmp;
 9    int cnt;
10    int a;
11    while (scanf("%d", &n) == 1)
12    {
13        prev = -1;
14        cnt = 0;
15        a = n;
16        while (n--)
17        {
18            scanf("%d", &tmp);
19            if (tmp == prev)
20            {
21                if (tmp == 0)
22                {
23                    prev = 1;
24                }
25                else
26                {
27                    prev = 0;
28                }
29                ++cnt;
30                continue;
31            }
32            prev = tmp;
33        }
34        if (cnt > a/2)
35            cnt = a-cnt;
36        printf("%d\n", cnt);
37    }
38    return 0;
39}

The Circumference of the Circle

本題在ZOJ上題號是1090，在POJ上是2242。題目描述如下：

Description

To calculate the circumference of a circle seems to be an easy task - provided you know its diameter. But what if you don't?

You are given the cartesian coordinates of three non-collinear points in the plane.
Your job is to calculate the circumference of the unique circle that intersects all three points.

Input

The input will contain one or more test cases. Each test case consists of one line containing six real numbers x1,y1, x2,y2,x3,y3, representing the coordinates of the three points. The diameter of the circle determined by the three points will never exceed a million. Input is terminated by end of file.

Output

For each test case, print one line containing one real number telling the circumference of the circle determined by the three points. The circumference is to be printed accurately rounded to two decimals. The value of pi is approximately 3.141592653589793.

Sample Input

0.0 -0.5 0.5 0.0 0.0 0.5

0.0 0.0 0.0 1.0 1.0 1.0

5.0 5.0 5.0 7.0 4.0 6.0

0.0 0.0 -1.0 7.0 7.0 7.0

50.0 50.0 50.0 70.0 40.0 60.0

0.0 0.0 10.0 0.0 20.0 1.0

0.0 -500000.0 500000.0 0.0 0.0 500000.0

Sample Output

3.14

4.44

6.28

31.42

62.83

632.24

3141592.65

分析：本題是一道比較容易的題，具體就考察了幾個數學公式的使用。本題的關鍵是求出內接三角形的外接圓直徑。而在圓的內接三角形的性質中有這樣一條：三角形的任何兩邊的乘積的等于第三邊上的高于其外接圓直徑的乘積。這樣問題就轉化為求接三角形的某一邊上的高，在知道三角形三個頂點的情況下，求其面積應該是件容易事，求得面積后，高的問題也就迎刃而解。求面積時，由于本人較懶，用的是海倫公式：S = ，其中p = (a+b+c)/2，a，b，c分別為三角形的三個變長，S=0.5*c*h，即可求得h。a*b=h*d，那么直徑d也就出來了。具體代碼如下.

#include <stdio.h>

#include <stdlib.h>

#include <math.h>

int main(void)

{

       double x1, y1, x2, y2, x3, y3;

       double l1, l2, l3;

       double p;

       double h;

       double d;

       while (scanf("%lf%lf%lf%lf%lf%lf", &x1, &y1, &x2, &y2, &x3, &y3) == 6)

       {

              l1 = sqrt(pow(x1-x2, 2) + pow(y1-y2, 2));

              l2 = sqrt(pow(x1-x3, 2) + pow(y1-y3, 2));

              l3 = sqrt(pow(x2-x3, 2) + pow(y2-y3, 2));

              p = (l1 + l2 + l3)/2;

              h = sqrt(p*(p-l1)*(p-l2)*(p-l3))*2/l3;

              d = l1*l2/h;

              printf("%.2f\n", 3.141592653589793*d);

       }

       return 0;

}