内容：

One day, as Sherlock Holmes was tracking down one very important criminal, he found a wonderful painting on the wall. This wall could be represented as a plane. The painting had several concentric circles that divided the wall into several parts. Some parts were painted red and all the other were painted blue. Besides, any two neighboring parts were painted different colors, that is, the red and the blue color were alternating, i. e. followed one after the other. The outer area of the wall (the area that lied outside all circles) was painted blue. Help Sherlock Holmes determine the total area of red parts of the wall.

Let us remind you that two circles are called concentric if their centers coincide. Several circles are called concentric if any two of them are concentric.  

Input

The first line contains the single integer n (1 ≤ n ≤ 100). The second line contains n space-separated integers ri (1 ≤ ri ≤ 1000) — the circles' radii. It is guaranteed that all circles are different.  

Output

Print the single real number — total area of the part of the wall that is painted red. The answer is accepted if absolute or relative error doesn't exceed 10 - 4.  

Examples
Input

1  
1

Output

3.1415926536

Input

3  
1 4 2

Output

40.8407044967

Note

In the first sample the picture is just one circle of radius 1. Inner part of the circle is painted red. The area of the red part equals π × 12 = π.

In the second sample there are three circles of radii 1, 4 and 2. Outside part of the second circle is painted blue. Part between the second and the third circles is painted red. Part between the first and the third is painted blue. And, finally, the inner part of the first circle is painted red. Overall there are two red parts: the ring between the second and the third circles and the inner part of the first circle. Total area of the red parts is equal (π × 42 - π × 22) + π × 12 = π × 12 + π = 13π

水题，但是因为我把一个‘ * ’号写成了‘ - ’号没发现导致wa了好几次，还以为是超时了，气死。。。

代码：

#include
#include
#include
#include
using namespace std;
int main()
{
int n;
cin>>n;
int s[n+5];
for(int i=0;i>s[i];
}
sort(s,s+n,greater());
double sum=0,p=3.141592653589;
if(n%2==0)
{
for(int i=0;i<n-1;i+=2)
{
sum+=p*(s[i]*s[i]-s[i+1]*s[i+1]);
}
printf("%.9lf\n",sum);
}
else{
for(int i=0;i<n-2;i+=2)
{
sum+=p*(s[i]*s[i]-s[i+1]*s[i+1]);//就是这里最后一个*号弄错了。。。
}
sum+=p*s[n-1]*s[n-1];
printf("%.9lf\n",sum);
}
}

内容：

Dr. Moriarty is about to send a message to Sherlock Holmes. He has a string s.

String p is called a substring of string s if you can read it starting from some position in the string s. For example, string "aba" has six substrings: "a", "b", "a", "ab", "ba", "aba".

Dr. Moriarty plans to take string s and cut out some substring from it, let's call it t. Then he needs to change the substring t zero or more times. As a result, he should obtain a fixed string u (which is the string that should be sent to Sherlock Holmes). One change is defined as making one of the following actions:

    Insert one letter to any end of the string.  
    Delete one letter from any end of the string.  
    Change one letter into any other one.

Moriarty is very smart and after he chooses some substring t, he always makes the minimal number of changes to obtain u.

Help Moriarty choose the best substring t from all substrings of the string s. The substring t should minimize the number of changes Moriarty should make to obtain the string u from it.  

Input

The first line contains a non-empty string s, consisting of lowercase Latin letters. The second line contains a non-empty string u, consisting of lowercase Latin letters. The lengths of both strings are in the range from 1 to 2000, inclusive.  

Output

Print the only integer — the minimum number of changes that Dr. Moriarty has to make with the string that you choose.  

Examples
Input

aaaaa  
aaa

Output

0

Input

abcabc  
bcd

Output

1

Input

abcdef  
klmnopq

Output

7

Note

In the first sample Moriarty can take any substring of length 3, and it will be equal to the required message u, so Moriarty won't have to make any changes.

In the second sample you should take a substring consisting of characters from second to fourth ("bca") or from fifth to sixth ("bc"). Then you will only have to make one change: to change or to add the last character.

In the third sample the initial string s doesn't contain any character that the message should contain, so, whatever string you choose, you will have to make at least 7 changes to obtain the required message.

题目大意：给出两个字符串s,u,通过将s中提取出的子字符串t 经过在任意一端删除字符，或在两端增加字符或者改变任意字符，使得t等于u；求最小操作数。

思路：顺序找到最长的匹配字符个数，再用len2减去，所得结果就是最少操作数。

代码：

#include
using namespace std;
int main()
{
string s,t;
while(cin>>s>>t)
{
int len1=s.length(),len2=t.length();
int ans=1e9,i,j,len=len2;
for(i=0;i<len1;i++)
{
int k,ct=0;
for(k=0,j=i;k<len2&&j<len1;k++,j++)
{
if(t[k]==s[j])
ct++;
}
ans=min(ans,len-ct);
}
for(i=0;i<len2;i++)
{
int k,ct=0;
for(k=0,j=i;k<len1&&j<len2;k++,j++)
{
if(t[j]==s[k])
ct++;
}
ans=min(ans,len-ct);
}
cout<<ans<<endl;
}
}

CodeForces - 157D

题目内容：

As Sherlock Holmes was investigating a crime, he identified n suspects. He knows for sure that exactly one of them committed the crime. To find out which one did it, the detective lines up the suspects and numbered them from 1 to n. After that, he asked each one: "Which one committed the crime?". Suspect number i answered either "The crime was committed by suspect number ai", or "Suspect number ai didn't commit the crime". Also, the suspect could say so about himself (ai = i).

Sherlock Holmes understood for sure that exactly m answers were the truth and all other answers were a lie. Now help him understand this: which suspect lied and which one told the truth?  

Input

The first line contains two integers n and m (1 ≤ n ≤ 105, 0 ≤ m ≤ n) — the total number of suspects and the number of suspects who told the truth. Next n lines contain the suspects' answers. The i-th line contains either "+ai" (without the quotes), if the suspect number i says that the crime was committed by suspect number ai, or "-ai" (without the quotes), if the suspect number i says that the suspect number ai didn't commit the crime (ai is an integer, 1 ≤ ai ≤ n).

It is guaranteed that at least one suspect exists, such that if he committed the crime, then exactly m people told the truth.  

Output

Print n lines. Line number i should contain "Truth" if suspect number i has told the truth for sure. Print "Lie" if the suspect number i lied for sure and print "Not defined" if he could lie and could tell the truth, too, depending on who committed the crime.  

Examples
Input

1 1  
+1

Output

Truth

Input

3 2  
-1  
-2  
-3

Output

Not defined  
Not defined  
Not defined

Input

4 1  
+2  
-3  
+4  
-1

Output

Lie  
Not defined  
Lie  
Not defined

Note

The first sample has the single person and he confesses to the crime, and Sherlock Holmes knows that one person is telling the truth. That means that this person is telling the truth.

In the second sample there are three suspects and each one denies his guilt. Sherlock Holmes knows that only two of them are telling the truth. Any one of them can be the criminal, so we don't know for any of them, whether this person is telling the truth or not.

In the third sample the second and the fourth suspect defend the first and the third one. But only one is telling the truth, thus, the first or the third one is the criminal. Both of them can be criminals, so the second and the fourth one can either be lying or telling the truth. The first and the third one are lying for sure as they are blaming the second and the fourth one.

题意：

题意：福尔摩斯正在处理一件案子。此时已经抓捕了n个嫌疑人，里面只可能有一个是真正的犯人。福尔摩斯正在审问这些嫌疑人。
每个嫌疑人的回答只有两种，一种表明他说编号为i的嫌疑人不是犯人，用-i表示；另一种表明他说编号为i的嫌疑人是犯人，
用+i表示。聪明的福尔摩斯已经知道了其中有m个人说的是真话。要求那些人说的是真话，那些人说的是假话
思路：

首先判断每个人是逃犯的情况符不符合m个人说实话
然后判断合适的情况有多少
判断一个人是不是逃犯只要判断他被人说是逃犯和不说他是逃犯的人和等于m就可以了，因为如果他是逃犯那么这些人都没说谎。
然后通过可能发生的的事情的数量判断每个人说没说谎
如果一个人说另一个人是逃犯而这个人不可能是逃犯那么就是说谎
如果一个人说另一个人是逃犯而这个人确实可能是逃犯那么这个人说的话就不被确定
如果一个人说另一个人是逃犯而可能的事实只有一种那么他却说的就是实话而这个人是对的
如果一个人说其他人不是逃犯而这个人可能是逃犯那么他说的话就是不确定的
如果一个人说另一个人不是逃犯而这个人不可能是逃犯那么他说的就是对的
如果一个人说另一个人不是逃犯而这个人可能是逃犯并且事实唯一那么他就说了谎那么个人确实是逃犯

#include
#include
#include
using namespace std;
int tu[100001], ky[100001], bky[100001], dui[100001],zgky=0,zgbky=0;
int main()
{
int n, m;
cin >> n >> m;
for (int a = 1; a <= n; a++) { scanf("%d", &tu[a]); if (tu[a] < 0)bky[-tu[a]]++,zgbky++; else ky[tu[a]]++, zgky++; } int kn = 0; for (int a = 1; a <= n; a++) { if (ky[a] + zgbky - bky[a] == m) { kn++; dui[a] = 1; } } for (int a = 1; a <= n; a++) { if (tu[a] > 0)
{
if (!dui[tu[a]])printf("Lie\n");
else if (kn > 1)printf("Not defined\n");
else printf("Truth\n");
}
else
{
if(!dui[-tu[a]])printf("Truth\n");
else if (kn > 1)printf("Not defined\n");
else printf("Lie\n");
}
}
}