bzoj2226

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

有点暴力。。。因为cout而re了两发。。

#include<bits/stdc++.h>
#define lowbit(a) ((a)&(-(a)))
#define clr(a,x) memset(a,x,sizeof(a))
#define rep(i,l,r) for(int i=l;i<(r);i++)
typedef long long ll;
using namespace std;
int read()
{
    char c=getchar();
    int ans=0,f=1;
    while(!isdigit(c)){
        if(c=='-') f=-1;
        c=getchar();
    }
    while(isdigit(c)){
        ans=ans*10+c-'0';
        c=getchar();
    }
    return ans*f;
}
const int maxn=1000009;
int n,cnt=0,phi[maxn],pri[maxn];
bool p[maxn];
void init(){
    clr(p,-1);
    rep(i,2,maxn){
        if(p[i]) pri[cnt++]=i,phi[i]=i-1;
        rep(j,0,cnt){
            if(pri[j]*i>=maxn) break;
            p[pri[j]*i]=0;
            if(i%pri[j]==0){
                phi[i*pri[j]]=phi[i]*pri[j];
                break;
            }else phi[i*pri[j]]=phi[i]*(pri[j]-1);
        }
    }
}
ll f(int x){
    if(x==1) return 1ll;
    return ll(phi[x])*ll(x)>>1;
}
int main()
{    
    init();
    int t=read();
    while(t--){
        n=read();
        ll ans=0;
        int i;
        for(i=1;i*i<n;i++){
            if(n%i==0){
                ans+=f(i)+f(n/i);
            }
        }
        if(i*i==n) ans+=f(i);
        printf("%lld
",ans*ll(n));
    }
    return 0;
}

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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