hdu-6434-欧拉函数

Problem I. Count

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 443    Accepted Submission(s): 232

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

 　　 ∑_i=1ⁿ∑^i-1_j=1 [gcd(i + j, i − j) = 1]

　　=  [gcd(2i-a,a)=1]    (1<=i<=n,1<=a<i)

　　=  [gcd(2i,a)=1] 　  (1<=i<=n,1<=a<i)

即对于每个 i, 求有多少个小于它的 a 满足 gcd(i, a) = 1 且a是奇数。

当i是偶数的时候贡献就是  phi(i) ,因为偶数与偶数一定不互质。

当i是奇数的时候贡献是phi(i)/2 ,因为i是奇数，假如x与i互质那么 i-x与i也互质，且x与i-x一个是奇数一个是偶数，所以数量是均等出现的。

 #include<bits/stdc++.h>
 using namespace std;
 #define LL long long
 #define mp make_pair
 #define pb push_back
 #define inf 0x7fffffffff
 #define pii pair<int,int>
 const int maxn=;
 vector<int>prime;
 bool is[maxn];
 LL f[maxn];
 void init(){
     f[]=;
     is[]=is[]=;
     for(int i=;i<=maxn;++i){
         if(!is[i]) prime.push_back(i),f[i]=i-;
         for(int j=;j<prime.size()&&1LL*i*prime[j]<=maxn;++j){
             is[i*prime[j]]=;
             if(i%prime[j]==){
                 f[i*prime[j]]=f[i]*prime[j];
                 break;
             }
             else{
                 f[i*prime[j]]=f[i]*(prime[j]-);
             }
         }
     }
     for(int i=;i<maxn;++i){
         if(i&)f[i]/=;
         f[i]+=f[i-];
     }
 }

 int main()
 {
     init();
     int t,n;
     scanf("%d",&t);
     while(t--){
         scanf("%d",&n);
         printf("%lld\n",f[n]);
     }
     return ;
 }