GCD
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Given the value of N, you will have to find the value of G. The definition of G is given below:
G =
i<N
∑
i=1
j
∑
≤N
j=i+1
GCD(i, j)
Here GCD(i, j) means the greatest common divisor of integer i and integer j.
For those who have trouble understanding summation notation, the meaning of G is given in the
following code:
G=0;
for(i=1;i<N;i++)
for(j=i+1;j<=N;j++)
{
G+=gcd(i,j);
}
/*Here gcd() is a function that finds
the greatest common divisor of the two
input numbers*/
Input
The input file contains at most 100 lines of inputs. Each line contains an integer N (1 < N < 4000001).
The meaning of N is given in the problem statement. Input is terminated by a line containing a single
zero.
Output
For each line of input produce one line of output. This line contains the value of G for the corresponding
N. The value of G will fit in a 64-bit signed integer.
Sample Input
10
100
200000
0
Sample Output
67
13015
G =
i<N
∑
i=1
j
∑
≤N
j=i+1
GCD(i, j)
Here GCD(i, j) means the greatest common divisor of integer i and integer j.
For those who have trouble understanding summation notation, the meaning of G is given in the
following code:
G=0;
for(i=1;i<N;i++)
for(j=i+1;j<=N;j++)
{
G+=gcd(i,j);
}
/*Here gcd() is a function that finds
the greatest common divisor of the two
input numbers*/
Input
The input file contains at most 100 lines of inputs. Each line contains an integer N (1 < N < 4000001).
The meaning of N is given in the problem statement. Input is terminated by a line containing a single
zero.
Output
For each line of input produce one line of output. This line contains the value of G for the corresponding
N. The value of G will fit in a 64-bit signed integer.
Sample Input
10
100
200000
0
Sample Output
67
13015
143295493160
1.设G(n)=gcd(1,n)+gcd(2,n)+……+gcd(n-1,n)。
gcd(x,n)=i是n的约数(x<n),按照这个约数进行分类。设满足gcd(x,n)=i的有ph(n/i)个,则有G(n)=sum(i*ph(n/i))。
ph(x)为欧拉函数。(代码下面有详细说明)
2.建立递推关系,G(n)=G(n-1)+gcd(1,n)+gcd(2,n)+……+gcd(n-1,n);
3.降低时间复杂度。用筛法预处理phi[x]表
用筛法预处理f(x)->枚举因数,更新其所有倍数求解
代码:
#include<cstdio>using namespace std;const int N=4000005;typedef long long ll;int ph[N];ll G[N];void init(){ for(int i=2;i<N;i++) ph[i]=i; for(int i=2;i<N;i++) { if(ph[i]==i) for(int j=i;j<N;j+=i) ph[j]=ph[j]/i*(i-1);//欧拉函数 for(int j=1;j*i<N;++j) G[j*i]+=(ll)j*ph[i]; } for(int i=3;i<N;i++) G[i]+=G[i-1];}int main(){ init(); int n; while(scanf("%d",&n),n) printf("%lld\n",G[n]); return 0;}欧拉函数:
在数论,对正整数n,欧拉函数是小于n的正整数中与n互质的数的数目(φ(1)=1)。此函数以其首名研究者欧拉命名(Euler'so totient function),它又称为Euler's totient function、φ函数、欧拉商数等。 例如φ(8)=4,因为1,3,5,7均和8互质。 从欧拉函数引伸出来在环论方面的事实和拉格朗日定理构成了欧拉定理的证明。
通式:
其中p1, p2……pn为x的所有质因数,x是不为0的整数。
φ(1)=1(唯一和1互质的数(小于等于1)就是1本身)。
注意:每种质因数只一个。 比如12=2*2*3那么φ(12)=12*(1-1/2)*(1-1/3)=4
若n是质数p的k次幂, ,因为除了p的倍数外,其他数都跟n互质。
设n为正整数,以 φ(n)表示不超过n且与n互素的正整数的个数,称为n的欧拉函数值
φ:N→N,n→φ(n)称为欧拉函数。
欧拉函数是积性函数——若m,n互质,
特殊性质:当n为奇数时, , 证明与上述类似。
若n为质数则
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