hdu.5203.Rikka with wood sticks(数学推导：一条长度为L的线段经分割后可以构成几种三角形）...

Rikka with wood sticks

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 600    Accepted Submission(s): 169

Problem Description

As we know, Rikka is poor at math. Yuta is worrying about this situation, so he gives Rikka some math tasks to practice. There is one of them:


Yuta have a wood stick of length  n which consists of n linked sticks of length 1. So it has n−1 connection points. Yuta finds that some sticks of length 1 of the wood stick are not strong. So he wants to choose three different connection points to cut it into four wood sticks and only one of them contains wood sticks which are not strong. And Yuta wants to minimize the length of this piece which contains bad wood sticks. Besides, Rikka wants to use the other three wood sticks to make a triangle. Now she wants to count the number of the ways to cut the wood sticks which can make both Yuta and herself happy.


It is too difficult for Rikka. Can you help her?

 

 

Input

This problem has multi test cases (no more than  20). For each test case, The first line contains two numbers n,m(1≤n≤1000000,1≤m≤1000). The next line contains m numbers (some of them may be same) – the position of each wood sticks which is not strong.

 

 

Output

For each test cases print only one number – the ways to cut the wood sticks.

 

 

Sample Input

6 1 3 5 1 3

 

 

Sample Output

2 0

 

 

Source

BestCoder Round #37 ($)

 

 1 #include<stdio.h>
 2 #include<algorithm>
 3 typedef __int64 ll ;
 4 int n , m ;
 5 int l , r ;
 6 
 7 bool ok (int a , int b , int c)
 8 {
 9     if (a + b > c && a + c > b && b + c > a)
10         return true ;
11     return false ;
12 }
13 
14 ll calc2 (int lg , int wd)
15 {
16     if (lg == wd) return 0 ;
17     if (lg < wd) std::swap (lg , wd) ;
18     int x = (lg - wd)/2 ;
19     if (x == 0) x ++ ;
20     if (ok (x , lg - x , wd))
21         return lg - 1 - (x - 1) * 2 ;
22     else
23         return lg - 1 - x * 2 ;
24 }
25 
26 ll calc1 (int lg)
27 {
28     ll tot = 0 ;
29     for (int i = 1 ; i <= lg/2  ; i++) {
30         tot += calc2 (i , lg - i) ;
31     }
32     return tot ;
33 }
34 
35 int main ()
36 {
37     //freopen ("a.txt" , "r" , stdin ) ;
38     while (~ scanf ("%d%d" , &n, &m)) {
39         l = n , r = 1 ;
40         for (int i = 0 ; i < m ; i++) {
41             int x ;
42             scanf ("%d" , &x) ;
43             l = std::min (l , x) ;
44             r = std::max (r , x) ;
45         }
46         if (l == 1) {
47             printf ("%I64d\n" , calc1 (n - r)) ;
48         }
49         else if (r == n) {
50             printf ("%I64d\n" , calc1 (l - 1)) ;
51         }
52         else printf ("%I64d\n" , calc2 (l - 1 , n - r)) ;
53     }
54     return 0 ;
55 }

View Code

 

转载于:https://www.cnblogs.com/get-an-AC-everyday/p/4425954.html