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(jiǎn)單但卻隱藏著陷阱，題目要求尋找的是最少的切換數(shù)，故從第二盞燈開始判斷處理得出的結(jié)論是不一定正確的。通過(guò)分析可以發(fā)現(xiàn)該題其實(shí)只存在兩種情況：奇數(shù)位置的燈開著或者偶數(shù)位置的燈開著。這樣可以直觀的處理該題：取奇數(shù)位置燈開著需要切換燈狀態(tài)數(shù)與偶數(shù)位置燈開著需切換燈狀態(tài)數(shù)的較小值。這樣的話需要掃描兩邊燈的狀態(tài)數(shù)組，開銷較大。進(jìn)一步分析，設(shè)a為奇數(shù)位置的燈開著需要切換的燈數(shù)，b為偶數(shù)位置燈開著需要切換的燈數(shù)。其實(shí)a+b=n。這樣本題就只需要掃描一遍數(shù)組，且進(jìn)一步優(yōu)化后存儲(chǔ)燈狀態(tài)的數(shù)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(hào)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(gè)數(shù)學(xué)公式的使用。本題的關(guān)鍵是求出內(nèi)接三角形的外接圓直徑。而在圓的內(nèi)接三角形的性質(zhì)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(wèn)題就轉(zhuǎn)化為求接三角形的某一邊上的高，在知道三角形三個(gè)頂點(diǎn)的情況下，求其面積應(yīng)該是件容易事，求得面積后，高的問(wèn)題也就迎刃而解。求面積時(shí)，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(gè)變長(zhǎng)，S=0.5*c*h，即可求得h。a*b=h*d，那么直徑d也就出來(lái)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;

}