SPOJ DCEPCA03

题目链接：·http://www.spoj.com/problems/DCEPCA03/

题目大意：欧拉函数为totient(n)，请你计算

H=0;
For (i=1; i<=n; i++) {
    For (j=1; j<=n; j++) {
        H = H + totient(i) * totient(j);
    }
}

解题思路：刚开始以为是什么神奇的性质，想了半天也没想出来。用筛法写出来后TLE了，然后发现，这**的就是求个前缀和然后O(n)一遍phi[i] * sum[n]啊。

想多了。。

代码：

int phi[maxn];
ll sum[maxn];
int n;

void dowork(){
    memset(phi, 0, sizeof(phi));
    phi[1] = 1;
    for(int i = 2; i <= 10000; i++){ 
        if(!phi[i]){ 
            for(int j = i; j <= 10000; j += i){
                if(!phi[j]) phi[j] = j;
                phi[j] = phi[j] / i * (i - 1);
            }
        }
    }
    memset(sum, 0, sizeof(sum));
    for(int i = 1; i <= 10000; i++) sum[i] = sum[i - 1] + phi[i];
}
void solve(){
    ll h = 0;
    for (int i = 1; i <= n; i++) {
        h += phi[i] * sum[n];
    }    
    printf("%lld
", h);
}
int main(){
    dowork();
    int t;
    scanf("%d", &t);
    while(t--){
        scanf("%d", &n);
        solve();
    }
}

题目：

DCEPCA03 - Totient Extreme

#math #totient

Given the value of N, you will have to find the value of H. The meaning of H is given in the following code:

H=0;
For (i=1; i<=n; i++) {
    For (j=1; j<=n; j++) {
        H = H + totient(i) * totient(j);
    }
}

Totient or phi function, φ(n) is an arithmetic function that counts the number of positive integers less than or equal to n that are relatively prime to n. That is, if n is a positive integer, then φ(n) is the number of integers k in the range 1 ≤ k ≤ n for which gcd(n, k) = 1

Constraints

0 < T <= 50
0 < N <= 10^4

Input

The first line contains T, the number of test cases. It is followed by T lines each containing a number N .

Output

For each line of input produce one line of output. This line contains the value of H for the corresponding N.

Example

Input:
2
3
10

Output:
16
1024