Sum Problem,简单题不能ac

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

int main()
{
int n;
while(scanf("%d",&n)==1)
{
int i;
int sum=0;
for(i=1;i<=n;i++)
sum+=i;
printf("%d",sum);
}
return 0;
}

printf("%d",sum);
->
printf("%d\n",sum);

这道题过不了有可能是楼上说的原因，还有一个注意点是数据规模有可能会爆掉，因此可以用long 类型来保存，另外最好不要用for循环来直接模拟，这样非常慢，有时间超限的危险，直接用等差数列公式就好
我的代码如下：

 #include\<iostream\>
using namespace std;
int main()
{
    int n,m;
    while(cin>>n)
    {
        if(n%2==0)
                {
                    m=n/2;
                    n++;
                }
        else 
                {
                    m=(n+1)/2;
                }
                cout<<n*m<<endl<<endl;
    }
    return 0;
}