C语言编程：1 求出1000以内所有偶数之和

Question

如题所述

Answer 1

#include<stdio.h>
int main()
{
printf("%d\n",(2+1000)*1000/2/2);
return 0;
}等差数列，直接输结果就是了

Answer 2

#include<stdio.h>
int main()
{
int i,sum=0;
for(i=2;i<=1000;i++)
if(i%2==0) sum+=i;
printf("sum=%d\n",sum);return 0;
}

Answer 3

#include<stdio.h>
void main()
{
int i;
float sum = 0;
for(i=2; i<=1000; i+=2)
sum += i;
printf("%.0f\n",sum);
}

Answer 4

#include<stdio.h>
int main()
{
int i;
float sum=0;
for(i=0;i<=1000;i++)
if(!(i%2))
sum+=i;
printf("%.0f\n",sum);
return 0;
}

Answer 5

#include<stdio.h>
int main()
{ int k=2;
float sum=0.0; //防止求和结果溢出
while ( k<=1000 )
{ sum=sum+i; i=i+2; }
printf("%f\n",sum);
system("pause");
}