531
备注:
|
← 于2008-02-23 15:36:54修订的的版本4 ⇥
808
converted to 1.6 markup
|
删除的内容标记成这样。 | 加入的内容标记成这样。 |
行号 1: | 行号 1: |
## page was renamed from 程序设计练习07——timus1068——Sum | |
行号 2: | 行号 3: |
http://acm.timus.ru/problem.aspx?space=1&num=1068 | |
行号 23: | 行号 25: |
http://acm.timus.ru/problem.aspx?space=1&num=1068 | |
行号 26: | 行号 28: |
{{{#!cplusplus #include <iostream> using namespace std; int main() { int n; cin >> n; if(n>0) { cout << n*(n+1)/2 << endl; } else { cout << n*(-n+1)/2+1 << endl; } } }}} |
Sum
http://acm.timus.ru/problem.aspx?space=1&num=1068
Time Limit: 2.0 second
Memory Limit: 1 000 K
Your task is to find the sum of all integer numbers lying between 1 and N inclusive.
1. Input
The input consists of a single integer N that is not greater than 10000 by it's absolute value.
2. Output
Write to the output file a single integer number that is the sum of all integer numbers lying between 1 and N inclusive.
3. Sample Input
{{{-3 }}}
4. Sample Output
{{{-5 }}}