UVA11424 GCD - Extreme (I)【欧拉函数打表】

Given the value of N, you will have to find the value of G. The definition of G is given below:
G=∑i=1i<N∑j=i+1j≤NGCD(i,j)G =\sum_{i=1}^{i<N}\sum_{j=i+1}^{j≤N}GCD(i, j)G=i=1∑i<Nj=i+1∑j≤NGCD(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 20000 lines of inputs. Each line contains an integer N (1 < N < 200001). 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
20000
0
Sample Output
67
13015
1153104356

问题链接：UVA11424 GCD - Extreme (I)
问题简述：（略）
问题分析：欧拉函数问题，不解释。
程序说明：（略）
参考链接：（略）
题记：（略）

AC的C++语言程序如下：

/* UVA11424 GCD - Extreme (I) */#include <bits/stdc++.h>using namespace std;/* 欧拉函数打表 */
const int N = 200001;
long long phi[N + 1];
void setphi()
{memset(phi, 0, sizeof(phi));phi[1] = 1;for(int i = 2; i <= N; i++)if(phi[i] == 0)for(int j = i; j <= N; j += i) {if(phi[j] == 0) phi[j] = j;phi[j] = phi[j] / i * (i - 1);}
}long long f[N], s[N];void init()
{memset(f, 0, sizeof f);for (int i = 1; i < N; i++)for (int j = 2 * i; j < N; j += i)f[j] += i * phi[j / i];s[2] = f[2];for (int i = 3; i < N; i++)s[i] = s[i - 1] + f[i];
}int main()
{setphi();init();int n;while (~scanf("%d", &n) && n)printf("%lld\n", s[n]);return 0;
}

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UVA11424 GCD - Extreme (I)【欧拉函数打表】

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