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.
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.
Print the minimum possible total tiredness if the friends meet in the same point.
3
4
1
101
99
2
5
10
9
找到中间数用求和公式算一下就行了
#include <iostream> #include <cstdio> #include <cmath> #include <cstring> #include <algorithm> #include <string> #include <vector> #include <queue> #include <stack> #include <set> #include <map> #define INF 0x3f3f3f3f #define lowbit(x) (x&(-x)) #define eps 0.00000001 #define pn printf(" ") 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 ", 1LL*(1+aa)*aa/2 + 1LL*(1+bb)*bb/2); }