版本2和3间的区别

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 }}}

   1 #include <iostream>
   2 using namespace std;
   3 
   4 int main()
   5 {
   6   int n;
   7   cin >> n;
   8   if(n>0) {
   9     cout << n*(n+1)/2 << endl;
  10   } else {
  11     cout << n*(-n+1)/2+1 << endl;
  12   }
  13 }

-  于2006-04-21 15:49:46修订的的版本2
  大小: 737
  编辑: czk
  备注:
+  于2006-04-21 20:30:44修订的的版本3
  大小: 808
  编辑: czk
  备注:
-删除的内容标记成这样。
+加入的内容标记成这样。
 行号 1:
+## page was renamed from 程序设计练习07——timus1068——Sum

"timus1068"版本比较

Sum

1. Input

2. Output

3. Sample Input

4. Sample Output