BNU 12846 LCM Extreme 最小公倍数之和（线性欧拉筛选+递推）

LCM Extreme

Time Limit: 3000ms

Memory 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

 /*
 题目大意：求lcm(1,2)+lcm(1,3)+lcm(2,3)+....+lcm(1,n)+....+lcm(n-2,n)+lcm(n-1,n)
 设sum(n)为sum(lcm(i,j))(1<=i<j<=n)之间最小公倍数的和,f(n)为sum(i*n/gcd(i,n))(1<=i<n)
 那么sum(n)=sum(n-1)+f(n)。可以用线性欧拉筛选+递推来做。
 */
 #include <iostream>
 #include <cstdio>
 #include <cstring>

 typedef unsigned long long LL;
 const int maxn=;
 LL phi[maxn],sum[maxn],f[maxn];

 void Euler()
 {
     memset(phi,,sizeof(phi));
     int i,j;phi[]=;
     for(i=;i<maxn;i++)
     {
         if(phi[i]) continue;
         for(j=i;j<maxn;j+=i)
         {
             if(!phi[j]) phi[j]=j;
             phi[j]=phi[j]/i*(i-);
         }
     }
     for(i=;i<maxn;i++) phi[i]=phi[i]*i/;//与i互质的数之和
 }

 void init()
 {
     Euler();
     memset(sum,,sizeof(sum));
     memset(f,,sizeof(f));
     int i,j;sum[]=f[]=;
     for(i=;i<maxn;i++)
     {
         f[i]+=phi[i]*i;//与i互质的数之间的lcm之和
         for(j=*i;j<maxn;j+=i)
             f[j]+=phi[i]*j;//gcd(x,j)=i的sum(lcm(x,j))
         sum[i]=sum[i-]+f[i];
     }
 }

 int main()
 {
     //freopen("in.txt","r",stdin);
     //freopen("out.txt","w",stdout);
     init();
     int t,icase=,n;
     scanf("%d",&t);
     while(t--)
     {
         scanf("%d",&n);
         printf("Case %d: %llu\n",++icase,sum[n]);
     }
     return ;
 }