hdu 2139 Calculate the formula

純數學題，推理如下：

已知:1^2+2^2+3^2+……+n^2 =n(n+1)(2n+1)/6 —①
那麼1^2+2^2+3^2+……+n^2+……+（2n+1）^2 =(2n+1)(n+1)(4n+3)/3 —②
又有2^2+4^2+6^2+……+(2n)^2 =4[1^2+2^2+3^2+……+n^2]=4*①=2n(n+1)(2n+1)/3 —③
設所求為S 比較②和③可知 S=②-③=(2n+1)(n+1)(4n+3)/3-2n(n+1)(2n+1)/3
=(2n+1)(n+1)(2n+3)/3 —④
因為S是2n+1項的和 把它一般化 則奇數項平方和一般公式Sn=n(n+1)(n+2)/6

注意要用long long保證不溢出！

#include<iostream>#include<vector>#include<map>#include<stack>#include<algorithm>#include<queue>#include<list>#include<set>#include<string.h>#include<stdlib.h>#include<math.h>#include<stdio.h>#include<ctype.h>#include<iomanip>using namespace std;#define LL long longint main(){    LL n;    while(scanf("%I64d",&n)!=EOF)    {        LL ans=n*(n+1)*(n+2)/6;        printf("%I64d\n",ans);    }    return 0;}