1 #include<stdio.h> 2 #include<math.h> 3 int main() 4 { 5 long long s2 , r2 , s3 , r3 , s4 , r4 , n ; 6 while( scanf("%lld" , &n) == 1 ) 7 { 8 s2 = n * ( n + 1 ) * ( 2 * n + 1) / 6 ; 9 s3 = n * n * ( n + 1 ) * ( n + 1 ) / 4 ; 10 s4 = n * ( n + 1 ) * (2 * n + 1) * ( 3 * n * n + 3 * n - 1) / 30 ; 11 12 r2 = n * n * ( n + 1 ) * ( n + 1 ) / 4 - s2; 13 r3 = n * n * n * (n+1) * (n+1) * (n+1) / 8 - s3; 14 r4 = n * n * n * n * (n+1) * (n+1) * (n+1) * (n+1) / 16 - s4; 15 printf("%lld %lld %lld %lld %lld %lld ", s2 , r2, s3, r3, s4 , r4) ; 16 } 17 return 0 ; 18 }
先分析二维情况：squares 数为 ∑n2 = n * ( n + 1 ) * ( 2 * n + 1) / 6
矩形总数为 ( n * (n+1) / 2)2
三维：squares 数为 ∑n3
矩形总数为 ( n * (n+1) / 2)3