bzoj2226

http://blog.csdn.net/benjaminpmlee/article/details/44947809

有点暴力。。。因为cout而re了两发。。

 1 #include<bits/stdc++.h>
 2 #define lowbit(a) ((a)&(-(a)))
 3 #define clr(a,x) memset(a,x,sizeof(a))
 4 #define rep(i,l,r) for(int i=l;i<(r);i++)
 5 typedef long long ll;
 6 using namespace std;
 7 int read()
 8 {
 9     char c=getchar();
10     int ans=0,f=1;
11     while(!isdigit(c)){
12         if(c=='-') f=-1;
13         c=getchar();
14     }
15     while(isdigit(c)){
16         ans=ans*10+c-'0';
17         c=getchar();
18     }
19     return ans*f;
20 }
21 const int maxn=1000009;
22 int n,cnt=0,phi[maxn],pri[maxn];
23 bool p[maxn];
24 void init(){
25     clr(p,-1);
26     rep(i,2,maxn){
27         if(p[i]) pri[cnt++]=i,phi[i]=i-1;
28         rep(j,0,cnt){
29             if(pri[j]*i>=maxn) break;
30             p[pri[j]*i]=0;
31             if(i%pri[j]==0){
32                 phi[i*pri[j]]=phi[i]*pri[j];
33                 break;
34             }else phi[i*pri[j]]=phi[i]*(pri[j]-1);
35         }
36     }
37 }
38 ll f(int x){
39     if(x==1) return 1ll;
40     return ll(phi[x])*ll(x)>>1;
41 }
42 int main()
43 {    
44     init();
45     int t=read();
46     while(t--){
47         n=read();
48         ll ans=0;
49         int i;
50         for(i=1;i*i<n;i++){
51             if(n%i==0){
52                 ans+=f(i)+f(n/i);
53             }
54         }
55         if(i*i==n) ans+=f(i);
56         printf("%lld
",ans*ll(n));
57     }
58     return 0;
59 }

View Code

2226: [Spoj 5971] LCMSum

Time Limit: 20 Sec Memory Limit: 259 MB
Submit: 889 Solved: 405
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Description

Given n, calculate the sum LCM(1,n) + LCM(2,n) + .. + LCM(n,n), where LCM(i,n) denotes the Least Common Multiple of the integers i and n.

Input

The first line contains T the number of test cases. Each of the next T lines contain an integer n.

Output

Output T lines, one for each test case, containing the required sum.

Sample Input

3
1
2
5

Sample Output

1
4
55

HINT

Constraints

1 <= T <= 300000
1 <= n <= 1000000

Source

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