hdu 5212 Code

最新推荐文章于 2018-02-01 15:43:00 发布

黎辰

最新推荐文章于 2018-02-01 15:43:00 发布

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文章标签： C++

本文链接：https://blog.youkuaiyun.com/qq_24451605/article/details/46398651

Code

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 456 Accepted Submission(s): 182

Problem Description

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)

For each case:

The first line contains an integer

N(1≤N≤10000).

The next line contains

N integers

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
题目大意：
易懂
题目分析：
因为数据范围很小，所以考虑数据范围内的每一个数作为gcd对于整个结果的贡献，
首先对于x，将它作为gcd的两个数一定都是它的倍数，而它的倍数有k个情况下，最多有k^2种情况将它作为最大公约数，
我们设f[x]是x作为最大公约数的数的对数，那么如果是两个是x倍数的数的最大公约数若不是x，那么一定是x的倍数，所以
f[x]+f[2x]+....+f[tx] = k*k , tx <= 10000 , k为x的倍数的个数
那么f[x]＝k*k - sigma[2-t](f[tx]),
所以先筛出每个数的倍数的个数，然后利用公式倒序求取之后，再枚举每一个数，求和即可
复杂度n*sqrt(n)

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#define MAX 10007
#define MOD 10007

using namespace std;

int f[MAX];
int n,a;
int cnt[MAX];

int main ( )
{
    while ( ~scanf ( "%d" , &n ) )
    {
        memset ( cnt , 0 , sizeof ( cnt ) );
        for ( int i = 1 ; i <= n ; i++ )
        {
            scanf ( "%d" , &a );
            for ( int j = 1 ; j*j <= a ; j++ )
                if ( a%j == 0 )
                {
                    cnt[j]++;
                    if ( a/j != j )
                        cnt[a/j]++;
                }
        }
        int ans = 0;
        for ( int x = 10000 ; x > 0 ; x-- )
        {
            f[x] = cnt[x]*cnt[x]%MOD;
            for ( int i = x*2 ; i <= 10000 ; i += x )
                f[x] = ( f[x] - f[i] + MOD) %MOD;
            int p = x*(x-1)%MOD;
            ans = ( ans + p*f[x]%MOD )%MOD;
        }
        printf ( "%d\n" , ans );
    }
    return 0;
}