On a Friday afternoon, some members of MaratonIME decided to watch a movie at CinIME.
There were n members who received popcorn buckets numbered from 1 to n.
At a certain moment, bucket 1 had one popcorn, bucket 2 had two popcorns and so on until bucket n, which had n popcorns. As good competitive programmers, they always prefer to simplify things, and decided to gather all the popcorn in just one bucket.
They proceeded on the following way: In bucket 2, they gather the popcorn from buckets 1 and 2. Then, in bucket 3, those of bucket 2 and 3 and so on until the last bucket. Formally, they perform n - 1 movements, on the i-th movement they join the popcorn from buckets i and i + 1 on bucket i + 1. However, they are known to be clumsy and at each moment they join two buckets, they let a single popcorn fall to the ground, which they promptly throw in the trash.
Jiang, the Sharp, realized that maybe the last bucket would be too small to hold all of the popcorn. Therefore, he asked for your help to determine how much popcorn should remain in the last bucket.
Given n, the number of members who decided to watch the movie, print the amount of popcorn that would remain in bucket n. Keep in mind that exactly one popcorn is lost at each step.
Input
The first line contains the integer n (2 ≤ n ≤ 3 * 109) – the number of members from MaratonIME who decided to watch the movie.
Output
An integer: The amount of popcorn the last bucket should have.
Examples
Input
2
Output
2
Input
3
Output
4
Input
1000000
Output
499999500001
题目大意：给你一个数n，让你求1~n的和减去n-1；
思路：这个题T了好几次。。一直都是直接用暴力来做，总是T，最后换成了等差数列AC了。
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main()
{
long long int sum, n;
scanf("%lld", &n);
sum = (n * (n - 1)) / 2 + 1;
printf("%lld\n", sum);
return 0;
}