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=20000010;
vector<int>prime;
bool is[maxn];
LL f[maxn];
void init(){
    f[1]=1;
    is[0]=is[1]=1;
    for(int i=2;i<=maxn;++i){
        if(!is[i]) prime.push_back(i),f[i]=i-1;
        for(int j=0;j<prime.size()&&1LL*i*prime[j]<=maxn;++j){
            is[i*prime[j]]=1;
            if(i%prime[j]==0){
                f[i*prime[j]]=f[i]*prime[j];
                break;
            }
            else{
                f[i*prime[j]]=f[i]*(prime[j]-1);
            }
        }
    }
    for(int i=1;i<maxn;++i){
        if(i&1)f[i]/=2;
        f[i]+=f[i-1];
    }
}


int main()  
{
    init();
    int t,n;
    scanf("%d",&t);
    while(t--){
        scanf("%d",&n);
        printf("%lld
",f[n]);
    }
    return 0;  
}