数字逻辑基础与verilog设计答案_数论中未解决的问题

题目链接
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

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]); } }