hdu 5212（容斥原理）

Code

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 77    Accepted Submission(s): 27


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

 

Sample Output

64

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

 

Source

BestCoder Round #39 ($)

 

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给你一个a数组，让你计算那段代码的结果，用容斥原理就ok，莫比乌斯反演也行，不过不会莫比乌斯反演（也是容斥原理）orz。。。改天学习下。

代码如下：

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<vector>
#include<cstring>
#include<map>
using namespace std;
typedef long long ll;
const int maxn = 1e4 + 10;
const ll mod = 10007;
#define rep(i,a,b) for(int i=(a);i<(b);i++)
#define pb push_back
int cnt[maxn],a[maxn];
ll d[maxn];
int main()
{
int n;
while(~scanf("%d",&n)) {
int Mgcd = 1;
for(int i = 0; i < n; i++) {
int x;
scanf("%d",&x);
Mgcd = max(Mgcd,x);
a[i] = x;
}
memset(cnt,0,sizeof(cnt[0])*Mgcd+20);
for(int i = 0; i < n; i++)cnt[a[i]]++;
ll ans = 0,res = 0;
/*d[i]表示任取两数gcd为i的方案数*/
for(ll i = Mgcd; i > 0; i--) {
ll tot = 0;
for(ll j = i; j <= Mgcd; j+= i)
{
tot += cnt[j];
d[i] = (d[i]-d[j])%mod;/*减去gcd为i的倍数的方案数(容斥原理)*/
}
/*gcd为i的方法数等于任选两个数(可重)是i的倍数的方案
除去gcd为i的倍数的情况(前面已经减掉了)*/
d[i] = (d[i] + tot*tot)%mod;
ans = (ans+d[i]*i*i)%mod;
res = (res+i*d[i])%mod;
}
cout<<((ans-res)%mod+mod)%mod<<endl;
}
return 0;
}