hdu 6434 ( Problem I. Count )

Posted 2020-12-31 tingtin

 

 

Problem I. Count

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 418    Accepted Submission(s): 216

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

 

 

 

 

 

令a=i-j,所以i+j=2*i-a，那么原式可以化简为：

a一定是奇数。

 

 

1 欧拉函数的几个性质：f(n) :欧拉函数值
2 1.若a为质数,phi[a]=a-1;
3 2.若a为质数,b mod a=0,phi[a*b]=phi[b]*a
4 3.若a,b互质,phi[a*b]=phi[a]*phi[b](当a为质数时,if b mod a!=0 ,phi[a*b]=phi[a]*phi[b])
5 4.gcd(kx,ky)=gcd(x,y)==1
6 5.gcd(kx-y,y)=gcd(kx,y)=gcd(x,y)==1
7 6.若N为奇数，那么f(2*n)=2*f(n)   
8 7.若n是质数p的k次幂，f(n)=p^k-p^(k-1)=(p-1)*p^(k-1)
9 因为除了p的倍数外，其他数都跟n互质。:1~n 每隔p个出现一个p的倍数

 

 

 

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<algorithm>
 4 #include<cstring>
 5 #include<cmath>
 6 #include<vector>
 7 #define N 20000009
 8 using namespace std;
 9 #define ll long long 
10 int pre[N],phi[N];
11 ll sum[N];
12 bool vis[N];
13 int t,n;
14 //欧拉函数 ：phi[i] j:(1~i)里有phi[i]个数令gcd(i,j)=1
15 void init()
16 {
17     pre[1]=1;
18     int i,j;
19     int pree=0;
20     for(i=2;i<=N;i++)
21     {
22         if(!vis[i]){
23             pre[++pree]=i;
24             phi[i]=i-1;
25         }
26         for(j=1;j<=pree&&i*pre[j]<=N;j++){
27             vis[i*pre[j]]=1;
28             if(i%pre[j]==0){
29                 phi[i*pre[j]]=phi[i]*pre[j];
30                 break;
31             }
32             else{
33                 phi[i*pre[j]]=phi[i]*phi[pre[j]];
34             }
35         }
36     }
37     for(int i=1;i<=N;i++){
38         if(i&1) phi[i]>>=1;          
39         sum[i]=sum[i-1]+phi[i];//求个前缀和即可
40         //if(i&1) phi[i]>>=1;
41         //因为a一定是奇数，那么偶数不变，因为偶数与偶数的gcd一定！=1
42         //奇数减半 如 7 ： 1 2 3 4 5 6  而只有1 3 5 有效
43     }
44 }
45 int main()
46 {  
47     init();
48     scanf("%d",&t);
49     while(t--){
50         scanf("%d",&n);
51         printf("%lld
",sum[n]);
52     }
53     return 0;
54 }