BNU 12846 LCM Extreme 最小公倍數之和（線性歐拉篩選+遞推）

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LCM ExtremeTime Limit: 3000msMemory Limit: 131072KB This problem will be judged on UVALive. Original ID: 5964
64-bit integer IO format:  %lld      Java class name:  Main

Find the result of the following code:
unsigned long long allPairLcm(int n){
unsigned long long res = 0;
for( int i = 1; i<=n;i++)
for(int j=i+1;j<=n;j++)
res += lcm(i, j);// lcm means least common multiple
return res;
}
A straight forward implementation of the code may time out.
Input
Input starts with an integer T (≤ 25000), denoting the number of test cases.
Each case starts with a line containing an integer n (1 ≤ n ≤ 5*106
).
Output
For each case, print the case number and the value returned by the function ‘allPairLcm(n)‘. As the
result can be large, we want the result modulo 2
64
.
Sample Input Output for Sample Input
4
2
10
13
100000
Case 1: 2
Case 2: 1036
Case 3: 3111
Case 4: 9134672774499923824

 1 /* 2 題目大意：求lcm(1,2)+lcm(1,3)+lcm(2,3)+....+lcm(1,n)+....+lcm(n-2,n)+lcm(n-1,n) 3 設sum(n)為sum(lcm(i,j))(1<=i<j<=n)之間最小公倍數的和,f(n)為sum(i*n/gcd(i,n))(1<=i<n) 4 那麼sum(n)=sum(n-1)+f(n)。可以用線性歐拉篩選+遞推來做。 5 */ 6 #include <iostream> 7 #include <cstdio> 8 #include <cstring> 9 10 typedef unsigned long long LL;11 const int maxn=5000005;12 LL phi[maxn],sum[maxn],f[maxn];13 14 void Euler()15 {16     memset(phi,0,sizeof(phi));17     int i,j;phi[1]=1;18     for(i=2;i<maxn;i++)19     {20         if(phi[i]) continue;21         for(j=i;j<maxn;j+=i)22         {23             if(!phi[j]) phi[j]=j;24             phi[j]=phi[j]/i*(i-1);25         }26     }27     for(i=1;i<maxn;i++) phi[i]=phi[i]*i/2;//與i互質的數之和28 }29 30 void init()31 {32     Euler();33     memset(sum,0,sizeof(sum));34     memset(f,0,sizeof(f));35     int i,j;sum[1]=f[1]=0;36     for(i=2;i<maxn;i++)37     {38         f[i]+=phi[i]*i;//與i互質的數之間的lcm之和39         for(j=2*i;j<maxn;j+=i)40             f[j]+=phi[i]*j;//gcd(x,j)=i的sum(lcm(x,j))41         sum[i]=sum[i-1]+f[i];42     }43 }44 45 int main()46 {47     //freopen("in.txt","r",stdin);48     //freopen("out.txt","w",stdout);49     init();50     int t,icase=0,n;51     scanf("%d",&t);52     while(t--)53     {54         scanf("%d",&n);55         printf("Case %d: %llu\n",++icase,sum[n]);56     }57     return 0;58 }