Code HDU - 5212 容斥

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#容斥

本文提供了一道来自HDU在线评测系统的题目解析，该题要求实现一个计算特定数学表达式的函数。文章详细展示了通过记录每个整数因子出现次数，并利用这些信息来高效计算所有可能的两两整数对的最大公约数及其相关表达式的值。

HDU - 5212

http://acm.hdu.edu.cn/showproblem.php?pid=5212

WLD likes playing with codes.One day he is writing a function.Howerver,his computer breaks down because the function is too powerful.He is very sad.Can you help him?

The function:

int calc
{

  int res=0;

  for(int i=1;i<=n;i++)

    for(int j=1;j<=n;j++)

    {

      res+=gcd(a[i],a[j])*(gcd(a[i],a[j])-1);

      res%=10007;

    }

  return res;

}

Input

There are Multiple Cases.(At MOST 10 10)

For each case:

The first line contains an integer N(1≤N≤10000) N(1≤N≤10000).

The next line contains N N integers a1,a2,...,aN(1≤ai≤10000) a1,a2,...,aN(1≤ai≤10000).

Output

For each case:

Print an integer,denoting what the function returns.

Sample Input

5
1 3 4 2 4

Sample Output

Hint

gcd(x,y) means the greatest common divisor of x and y.

/**
 *               ii.                                         ;9ABH,
 *              SA391,                                    .r9GG35&G
 *              &#ii13Gh;                               i3X31i;:,rB1
 *              iMs,:,i5895,                         .5G91:,:;:s1:8A
 *               33::::,,;5G5,                     ,58Si,,:::,sHX;iH1
 *                Sr.,:;rs13BBX35hh11511h5Shhh5S3GAXS:.,,::,,1AG3i,GG
 *                .G51S511sr;;iiiishS8G89Shsrrsh59S;.,,,,,..5A85Si,h8
 *               :SB9s:,............................,,,.,,,SASh53h,1G.
 *            .r18S;..,,,,,,,,,,,,,,,,,,,,,,,,,,,,,....,,.1H315199,rX,
 *          ;S89s,..,,,,,,,,,,,,,,,,,,,,,,,....,,.......,,,;r1ShS8,;Xi
 *        i55s:.........,,,,,,,,,,,,,,,,.,,,......,.....,,....r9&5.:X1
 *       59;.....,.     .,,,,,,,,,,,...        .............,..:1;.:&s
 *      s8,..;53S5S3s.   .,,,,,,,.,..      i15S5h1:.........,,,..,,:99
 *      93.:39s:rSGB@A;  ..,,,,.....    .SG3hhh9G&BGi..,,,,,,,,,,,,.,83
 *      G5.G8  9#@@@@@X. .,,,,,,.....  iA9,.S&B###@@Mr...,,,,,,,,..,.;Xh
 *      Gs.X8 S@@@@@@@B:..,,,,,,,,,,. rA1 ,A@@@@@@@@@H:........,,,,,,.iX:
 *     ;9. ,8A#@@@@@@#5,.,,,,,,,,,... 9A. 8@@@@@@@@@@M;    ....,,,,,,,,S8
 *     X3    iS8XAHH8s.,,,,,,,,,,...,..58hH@@@@@@@@@Hs       ...,,,,,,,:Gs
 *    r8,        ,,,...,,,,,,,,,,.....  ,h8XABMMHX3r.          .,,,,,,,.rX:
 *   :9, .    .:,..,:;;;::,.,,,,,..          .,,.               ..,,,,,,.59
 *  .Si      ,:.i8HBMMMMMB&5,....                    .            .,,,,,.sMr
 *  SS       :: h@@@@@@@@@@#; .                     ...  .         ..,,,,iM5
 *  91  .    ;:.,1&@@@@@@MXs.                            .          .,,:,:&S
 *  hS ....  .:;,,,i3MMS1;..,..... .  .     ...                     ..,:,.99
 *  ,8; ..... .,:,..,8Ms:;,,,...                                     .,::.83
 *   s&: ....  .sS553B@@HX3s;,.    .,;13h.                            .:::&1
 *    SXr  .  ...;s3G99XA&X88Shss11155hi.                             ,;:h&,
 *     iH8:  . ..   ,;iiii;,::,,,,,.                                 .;irHA
 *      ,8X5;   .     .......                                       ,;iihS8Gi
 *         1831,                                                 .,;irrrrrs&@
 *           ;5A8r.                                            .:;iiiiirrss1H
 *             :X@H3s.......                                .,:;iii;iiiiirsrh
 *              r#h:;,...,,.. .,,:;;;;;:::,...              .:;;;;;;iiiirrss1
 *             ,M8 ..,....,.....,,::::::,,...         .     .,;;;iiiiiirss11h
 *             8B;.,,,,,,,.,.....          .           ..   .:;;;;iirrsss111h
 *            i@5,:::,,,,,,,,.... .                   . .:::;;;;;irrrss111111
 *            9Bi,:,,,,......                        ..r91;;;;;iirrsss1ss1111
 */
#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int maxx = 1010;
const int mod = 10007;

int gcd(int a,int b)
{
	return !b ? a : gcd(b,a%b); 
}



int a[20000];

int f[20000];
int factor[20000];
int main()
{
	int n;
	while(~scanf("%d",&n))
	{
		memset(factor,0,sizeof(factor));
		int maxn = 0;
		for(int i = 0;i < n;i++)//输入的时候把每个数的因子保存到factor[]; 
		{
			scanf("%d",&a[i]);
			maxn = max(maxn,a[i]);
			for(int j = 1;j*j <= a[i];j++)
			{
				if(!(a[i]%j))
				{
					factor[j]++;
					if((a[i]/j)!=j)
					{
						factor[a[i]/j]++;
					}
				}
			}
		}
		//f[i]为 公因子只是i的数 的对数(aka. cp)("对三，要不起 "的对，不是log)
		// 注意这个 只^ 字  
		//gcd是最大,所以 f[i] = factor[i]*factor[i]后要减去  公因子是i倍数的f[] 
		int ans = 0;
		for(int i = maxn;i >= 1;i--)
		{
			f[i] = factor[i]*factor[i]%mod;
			for(int j = i*2;j <= maxn;j+=i)//因为这个for 用来减 
			{
				f[i] = ((f[i]-f[j]+mod)%mod);
			}
			int now = i*(i-1)%mod;// gcd(a[i],a[j])*(gcd(a[i],a[j])-1)
			ans = ((ans + f[i]*now)%mod);// f[i]*now就是(gcd为i的cp的数量)*(  gcd(a[i],a[j])*(gcd(a[i],a[j])-1)   ) 
			
		}
		printf("%d\n",ans);
		
	 } 





 return 0;
}