Problem Description
You just need to calculate the sum of the formula: 1^2+3^2+5^2+……+ n ^2.
Input
In each case, there is an odd positive integer n.
Output
Print the sum. Make sure the sum will not exceed 2^31-1
Sample Input
3
Sample Output
10
为了C币。。。
long 一个 sum = 0;
这个还是for循环
for (int i = 0; i <= n; i += 2)
每次sum += i * i;
最后 printf