ACM-sum of all integer numbers
2015-05-27 12:01
441 查看
描述 Your task is to find the sum of all integer numbers lying between 1 and N inclusive.
输入There are multiple test cases.
The input consists of a single integer N that is not greater than 10000 by it's absolute value.输出Write a single integer number that is the sum of all integer numbers lying between 1 and N inclusive.样例输入
样例输出
代码：
输入There are multiple test cases.
The input consists of a single integer N that is not greater than 10000 by it's absolute value.输出Write a single integer number that is the sum of all integer numbers lying between 1 and N inclusive.样例输入
3
样例输出
6
代码：
01.
#include<iostream>
02.
using
namespace
std;
03.
#include <stdio.h>
04.
int
main()
05.
{06.
int
n;
07.
int
t;
08.
while
(
scanf
(
"%d"
,&n)==1)
09.
{10.
if
(n > 0)
11.
printf
(
"%d\n"
,(1+n)*n/2);
12.
else
if
(n ==0)
13.
printf
(
"%d\n"
,1);
14.
else
15.
{16.
t = -n;
17.
printf
(
"%d\n"
,-(1+t)*t/2+1);
18.
}
19.
}
20.
21.
return
0;
22.
}
内容来自用户分享和网络整理，不保证内容的准确性，如有侵权内容，可联系管理员处理 
相关文章推荐
- 南阳acm 436 sum of all integer numbers
- sum of all integer numbers
- NYOJ 436 sum of all integer numbers
- NYOJ 436 sum of all integer numbers（坑人，递归）
- sum of all integer numbers
- sum of all integer numbers
- Given an integer, return all sequences of numbers that sum to it
- NYOJ-sum of all integer numbers
- NYOJ - 436 - sum of all integer numbers(注意a<0)
- 南阳理工：sum of all integer numbers
- sum of all integer numbers
- NYOJ436 sum of all integer numbers
- NYOJ436 sum of all integer numbers
- 436 sum of all integer numbers
- sum of all integer numbers
- NYOJ 46 sum of all integer numbers
- sum of all integer numbers（坑题，水题）
- 北大ACM2739——Sum of Consecutive Prime Numbers
- Problem 21 of Evaluate the sum of all the amicable numbers under 10000.
- 北大ACM 2739 Sum of Consecutive Prime Numbers