Sum Problem
Time Limit: 1000/500 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 609664 Accepted Submission(s): 154237
Problem Description
Hey, welcome to HDOJ(Hangzhou Dianzi University Online Judge).
In this problem, your task is to calculate SUM(n) = 1 + 2 + 3 + … + n.
Input
The input will consist of a series of integers n, one integer per line.
Output
For each case, output SUM(n) in one line, followed by a blank line. You may assume the result will be in the range of 32-bit signed integer.
Sample Input
1
100
Sample Output
1
5050
求和
输入n,求1+2+3+…+n
直接用求和公式
要注意的是输入的数能否被2整除,因为整型会自动去掉小数点后的数
还有最后需要两个回车
仔细看题目的输出
1和5050中有一个空行
#include
int main(void)
{
int a, sum;
while(~scanf("%d", &a))
{
if(a % 2 == 0)sum = a/2*(a + 1);
else sum = (a + 1)/2*a;
printf("%dnn",sum);
}
return 0;
}
标签：oj,sum,杭电,Limit,Others,integer,Problem,line,1001
来源： https://blog.csdn.net/FaceSeace/article/details/87951019