Codeforces Round #277 (Div. 2)---A. Calculating Function (规律)

时间:2023-03-09 16:04:48
Codeforces Round #277 (Div. 2)---A. Calculating Function   (规律)
time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

For a positive integer n let's define a function f:

f(n) =  - 1 + 2 - 3 + .. + ( - 1)nn

Your task is to calculate f(n) for a given integer n.

Input

The single line contains the positive integer n (1 ≤ n ≤ 1015).

Output

Print f(n) in a single line.

Sample test(s)
input
4
output
2
input
5
output
-3
Note

f(4) =  - 1 + 2 - 3 + 4 = 2

f(5) =  - 1 + 2 - 3 + 4 - 5 =  - 3

解题思路:大水题一枚。直接找规律。n%2==0时,f = n/2; 否则,f = -(n+1)/2.

AC代码:

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;
#define INF 0x7fffffff int main()
{
// #ifdef sxk
// freopen("in.txt","r",stdin);
// #endif
long long n;
while(scanf("%lld",&n)!=EOF)
{
if(n & 1) printf("%lld\n", -(n+1)/2);
else printf("%lld\n", n/2);
}
return 0;
}