1068. Sum
Time Limit: 2.0 second
Memory Limit: 16 MB

Your task is to find the sum of all integer numbers lying between 1 and N inclusive.
Input
The input consists of a single integer N that is not greater than 10000 by it's absolute value.
Output
Write a single integer number that is the sum of all integer numbers lying between 1 and N inclusive.
 

#include "stdio.h"
int main()
{
   int n,sum = 0;
   int i;
   scanf("%d",&n);
   if(n > 10000)
       return 0;
   if(n < 0)
   {
     for( i = n; i <=0;i ++)
     {
       sum += i;
     }
     sum += 1;
   }
   else
   {
      for(i = 0;i <= n;i ++)
      {
         sum += i;
      }
   }
   printf("%d\n",sum);
   return 0;
}
//////////
#include "stdio.h"
#include "math.h"
int main()
{
   int n,sum = 0;
   int i;
   scanf("%d",&n);
   if(n > 10000)
       return 0;
   printf("%d\n",(n + 1) * (abs(n - 1) + 1)/2);
   return 0;
}
上面是一个acm简单题的两个不同的程序，测试数据相同 结果相同，为什么上面一个通不过，下面一个却可以 。。。

好像上一个缺少#include "math.h"

哪里缺？第一个不需要的吧

 if(n < 0)
   {
     for( i = n; i <=0;i ++)
     {
       sum += i;
     }
     sum += 1;
   }
   else
   {
      for(i = 0;i <= n;i ++)
      {
         sum += i;
      }
   }

在此代码外围加个else{},仅供参考！！！

两个都可以  运行没有什么不对

·····这个不用math.h也是可以的·