Summer Training day4 Code hdu5212 莫比乌斯反演

最新推荐文章于 2019-03-17 19:57:47 发布

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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 1010)

For each case:

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

The next line contains NN 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.

方法1：莫比乌斯反演

统计1-10000中每个数出现的次数并存在num数组里面

从先枚举i和j变成先枚举gcd

其中sum数组表示的含义是

sum(i) = num(i)+num(2i)+...+num(ni)

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
typedef long long LL;
const int MAX = 10007;
const int MOD = 10007;
int nums[MAX];
int sums[MAX];
const int maxn = 222232;  
int mu[maxn],a[maxn],prime[maxn],cnt[maxn],num[maxn];  
bool vis[maxn];  
int n,pnum;  
int aa,b,c,d,k;
void mobeius(int N)  
{  
    pnum=0;  
    vis[1]=mu[1]=1;  
    for(int i=2;i<=N;i++)  
    {  
        if(!vis[i])  
        {  
            mu[i]=-1;  
            prime[pnum++]=i;  
        }  
        for(int j=0;j<pnum;j++)  
        {  
            if(i*prime[j]>N)break;  
            vis[i*prime[j]]=1;//É¸µôºÏÊý  
            if(i%prime[j]==0)  
            {  
                mu[i*prime[j]]=0;  
                break;//±£Ö¤ºÏÊýÊ¹ÓÃ×îÐ¡µÄËØÊýÉ¸µôµÄ  
            }  
            mu[i*prime[j]]=-mu[i];  
        }  
    }  
}  
int main(){
	mobeius(MAX);
	int N;
	while(scanf("%d",&N) != EOF){
		memset(nums,0,sizeof(nums));
		memset(sums,0,sizeof(sums));
		for(int i = 0 ;i < N;i++){
			int tmp;
			scanf("%d",&tmp);
			nums[tmp] ++;
		}
		for(int i = 1;i <= 10000;i++){
			for(int j = i;j <= 10000;j += i){
				sums[i] += nums[j];
			}
			sums[i] %= MOD;
		}
		LL ans = 0;
		for(int n = 1;n <= 10000;n++){
			LL g = 0;
			for(int j = n;j <= 10000;j += n){
				g = (g + mu[j/n] * sums[j] % MOD * sums[j]) % MOD;
			}
			ans = (ans + g * n % MOD * (n-1)) % MOD;
		}
		printf("%lld\n",ans);
	}
	return 0;
}

方法2：

依然同方法一一样处理，只不过在到达这一步以后我们可以方便的求出后面一半的值

即这个式子的值

我们定义上面的式子为F[k]

那么我们如果求F的时候将k从大到小递减，那么F[i] = sum(i)*sum(i) - F[2i] -F[3i] -.....

这里sum(i)的定义与方法一中的定义一致

代码：

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
#define int long long
const int MOD = 10007;
const int MAX = 10006;
int N;
int num[MAX];
int F[MAX];
main(){
	while(scanf("%lld",&N) != EOF){
		int ma = 0;
		memset(num,0,sizeof(num));
		memset(F,0,sizeof(F));
		for(int i = 0;i < N;i++){
			int tmp;scanf("%lld",&tmp);
			num[tmp] ++;
			ma = max(ma,tmp);
		}	
		int ans = 0;
		for(int i = ma;i >= 1;i--){
			int tmp = 0;
			int sum = 0;
			for(int j = i;j <= ma;j += i){
				sum = (sum + num[j]) % MOD;
				tmp = (F[j] + tmp) % MOD;
			}
			F[i] = (sum * sum % MOD - tmp + MOD )% MOD;
			ans += F[i] * i % MOD * (i-1) % MOD;
		}	
		printf("%lld\n",ans % MOD);
	}
	return 0;
}