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Problem Description
Multiple query, for each n, you need to get
n i-1
∑ ∑ [gcd(i + j, i - j) = 1]
i=1 j=1

Input
On the first line, there is a positive integer T, which describe the number of queries. Next there are T lines, each line give a positive integer n, as mentioned above.
T<=1e5, n<=2e7

Output
Your output should include T lines, for each line, output the answer for the corre- sponding n.

Sample Input
4
978
438
233
666

Sample Output
194041
38951
11065
89963

Source
2018 Multi-University Training Contest 10
思路：
变换差不多就是这样，于是对于每一个i就变成了求2i的欧拉函数除2了，求一下前缀和就行。

#include<bits/stdc++.h>using namespace std;typedef long long ll;const int N=2e7+5;ll euler[N<<1],sum[N],prim[N<<1];void get_euler(int n){    //欧拉函数    memset(euler, 0, sizeof euler);  euler[1] = 1;  int id = 0;  for(int i = 2; i < n; ++i) {       if(!euler[i]) {         euler[i] = i - 1;      prim[id++] = i;    }    for(int j = 0; j < id && prim[j]*i <n; ++j) {         if(i % prim[j]) {           euler[i*prim[j]] = euler[i] * (prim[j]-1);      } else {           euler[i*prim[j]] = euler[i] * prim[j];        break;      }    }  }}int main(){       int n,T;    get_euler(N<<1);    sum[1]=0;    for(int i=2;i<=N;++i) sum[i]=euler[2*i]/2,sum[i]+=sum[i-1];    scanf("%d",&T);    while(T--)    {           scanf("%d",&n);        printf("%lld\n",sum[n]);    }}

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