Project Euler Problem 6

Problem 6

区間[1, n]の各整数の二乗の総和 = n(n+1)(2n+1)/6
区間[1, n]の各整数の総和 = n(n+1)/2

求める答えは|(n(n+1)/2)^2 - n(n+1)(2n+1)/6|となる。

Scheme

(use srfi-1)
(use numbers)

(define (square-of-sum n)
  (let ((sum (* n (+ n 1) 1/2)))
    (* sum sum)))

(define (sum-of-squares n)  
  (* n (+ n 1) (+ (* 2 n) 1) 1/6))

(define (p006)
  (print (- (square-of-sum 100) (sum-of-squares 100))))

(p006)

C++

#include <cstdio>

int square_of_sum(int n)
{
    int t = n*(n+1)/2;
    return t*t;
}

int sum_of_squares(int n)
{
    return n*(n+1)*(2*n+1)/6;
}

int main()
{
    printf("%d\n", square_of_sum(100)-sum_of_squares(100));
    return 0;
}