CodeForces 251A 2점

Little Petya likes points a lot. Recently his mom has presented him n points lying on the line OX. Now Petya is wondering in how many ways he can choose three distinct points so that the distance between the two farthest of them doesn't exceed d.
Note that the order of the points inside the group of three chosen points doesn't matter.
Input
The first line contains two integers: n and d (1 ≤ n ≤ 105; 1 ≤ d ≤ 109). The next line contains n integers x1, x2, ..., xn, their absolute value doesn't exceed 109 — the x-coordinates of the points that Petya has got.
It is guaranteed that the coordinates of the points in the input strictly increase.
Output
Print a single integer — the number of groups of three points, where the distance between two farthest points doesn't exceed d.
Please do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64dspecifier.
Sample Input
Input

4 3
1 2 3 4

Output

4

Input

4 2
-3 -2 -1 0

Output

2

Input

5 19
1 10 20 30 50

Output

1

Hint
In the first sample any group of three points meets our conditions.
In the seconds sample only 2 groups of three points meet our conditions: {-3, -2, -1} and {-2, -1, 0}.
In the third sample only one group does: {1, 10, 20}.
제목: n개점, 거리 d, 집합 중 세 개마다 가장 멀고 먼 거리가 d를 초과하지 않는 집합을 찾아낸다.
사고방식: 출발점을 선정한 다음에 2분으로 상계를 찾아 개수를 구하고 복잡도는 nlogn이다
코드:

#include <stdio.h>
#include <string.h>

long long n, d;
long long i, j;
long long num[100005];
long long ans = 0;

long long find(long long  start, long long i, long long j) {
	long long mid;
	while (i < j) {
		mid = (i + j) / 2;
		if (num[mid]  - num[start] <= d) {
			i = mid + 1;
		}
		else {
			j = mid;
		}
	}
	mid = (i + j) / 2;
	if (num[mid] - num[start] > d)
		mid --;
	return mid;
}


int main() {
	scanf("%lld%lld", &n, &d);
	for (i = 0; i < n; i ++)
		scanf("%lld", &num[i]);
	for (i = 0; i < n - 2 ; i ++) {
		int j = find(i, i + 2, n - 1);
		if (j - i < 2) continue;
		ans += (j - i) * (j - i - 1) / 2;
	}
	printf("%lld
", ans);
	return 0;
}