UVA - 11426 GCD - Extreme (II)

Given the value of N, you willhave to find the value of G. The definition of G is given below:

 

Here GCD(i,j) means the greatest common divisor of integer i and integer j.

 

For those who have troubleunderstanding summation notation, the meaning of G is given in the followingcode:

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 100lines of inputs. Each line contains an integer N (1<N<4000001). Themeaning of N is given in the problem statement. Input is terminated by a linecontaining a single zero. 

 

Output

For each lineof input produce one line of output. This line contains the value of G for thecorresponding N. The value of G will fit in a 64-bit signed integer.

 

Sample Input                              Output for SampleInput

10

100

200000

0

 

67

13015

143295493160

 

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Problemsetter: Shahriar Manzoor

SpecialThanks: Syed Monowar Hossain

题意：输入正整数n,求gcd(1, 2)+gcd(1, 3)+gcd(2, 3)+...+gcd(n-1, n),即所有满足1<=i<j<=n的数对(i, j)所对应的gcd(i,j)之和。

思路：设f(n) = gcd(1, n)+gcd(2, n) + ...+gcd(n-1, n),则答案就为S（n） = f(2)+f(3)+...+f(n)，只需求出f(n),就可以递推出答案：S(n) = S(n-1) + f(n)，

我们再用g(n,i)表示满足gcd(x, n)=i且x<n的正整数x的个数，则f(n)=sum{i*g(n, i)}， 注意到gcd(x, n) = i的充要条件是gcd(x/i, n/i) = 1，因此满足条件的有x/i有phi(n/i)，

phi(a) 表示的是不超过x且和x互素的整数个数，我们通过类似筛选法的方法，每次去更新某个数i的倍数，而不是去枚举每个n的约数。

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>typedef long long ll;using namespace std;const int maxn = 4000001;int phi[maxn];ll S[maxn], f[maxn];void phi_table(int n) {for (int i = 2; i <= n; i++)phi[i] = 0;phi[1] = 1;for (int i = 2; i <= n; i++)if (!phi[i])for (int j = i; j <= n; j += i) {if (!phi[j])phi[j] = j;phi[j] = phi[j] / i * (i-1);}}int main() {phi_table(maxn);memset(f, 0, sizeof(f));for (int i = 1; i <= maxn; i++)for (int n = i*2; n <= maxn; n += i)f[n] += i * phi[n/i];S[2] = f[2];for (int n = 3; n <= maxn; n++)S[n] = S[n-1] + f[n];int n;while (scanf("%d", &n) != EOF && n) printf("%lld\n", S[n]);return 0;}