Codeforces Round #468 (Div. 2, based on Technocup 2018 Final Round)A. Friends Meeting-白红宇

Codeforces Round #468 (Div. 2, based on Technocup 2018 Final Round)A. Friends Meeting

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发布时间：2019-06-12

本文共 1803 字，大约阅读时间需要 6 分钟。

Two friends are on the coordinate axis Ox in points with integer coordinates. One of them is in the point x1 = a, another one is in the point x2 = b.

Each of the friends can move by one along the line in any direction unlimited number of times. When a friend moves, the tiredness of a friend changes according to the following rules: the first move increases the tiredness by 1, the second move increases the tiredness by 2, the third — by 3 and so on. For example, if a friend moves first to the left, then to the right (returning to the same point), and then again to the left his tiredness becomes equal to 1 + 2 + 3 = 6.

The friends want to meet in a integer point. Determine the minimum total tiredness they should gain, if they meet in the same point.

Input

The first line contains a single integer a (1 ≤ a ≤ 1000) — the initial position of the first friend.

The second line contains a single integer b (1 ≤ b ≤ 1000) — the initial position of the second friend.

It is guaranteed that a ≠ b.

Output

Print the minimum possible total tiredness if the friends meet in the same point.

Examples

input

3 4

output

input

101 99

output

input

5 10

output

9 找到中间数用求和公式算一下就行了

#include 
        
         #include 
         
          #include 
          
           #include 
           
            #include 
            
             #include 
             
              #include 
              
               #include 
               
                #include 
                
                 #include 
                 
                  #include 
                  #define INF 0x3f3f3f3f#define lowbit(x) (x&(-x))#define eps 0.00000001#define pn printf("\n")using namespace std;typedef long long ll;const int maxn = 1e5+7;int main(){ int a,b; scanf("%d%d",&a,&b); int mid = abs(a-b); int aa = mid/2 + mid%2, bb = mid/2; printf("%I64d\n", 1LL*(1+aa)*aa/2 + 1LL*(1+bb)*bb/2);}

转载于:https://www.cnblogs.com/HazelNut/p/8621331.html

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