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/*
Starting with the number 1 and moving to the right in a clockwise direction a 5 by 5 spiral is formed as follows:
21 22 23 24 25
20 7 8 9 10
19 6 1 2 11
18 5 4 3 12
17 16 15 14 13
It can be verified that the sum of the numbers on the diagonals is 101.
What is the sum of the numbers on the diagonals in a 1001 by 1001 spiral formed in the same way?
Ans:
Clearly top right corner is odd power
and all other corners are symetric around bottom right corner
so sum of corners in nth circle is (2n+1)^2 + 3( (2n+1)^2 - 2*(2n+1 -1))
2n+1=1001 n=500
sum = sigma 4*(4*n*n+n+1)
*/
#include<stdio.h>
int main()
{
unsigned int n,s=0;
for(n=1;n<=500;++n)
s+=4*(4*n*n+n+1);
printf("%u",s+1);
}