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=5000005;
LL phi[maxn],sum[maxn],f[maxn];

void Euler()
{
    memset(phi,0,sizeof(phi));
    int i,j;phi[1]=1;
    for(i=2;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-1);
        }
    }
    for(i=1;i<maxn;i++) phi[i]=phi[i]*i/2;//与i互质的数之和
}

void init()
{
    Euler();
    memset(sum,0,sizeof(sum));
    memset(f,0,sizeof(f));
    int i,j;sum[1]=f[1]=0;
    for(i=2;i<maxn;i++)
    {
        f[i]+=phi[i]*i;//与i互质的数之间的lcm之和
        for(j=2*i;j<maxn;j+=i)
            f[j]+=phi[i]*j;//gcd(x,j)=i的sum(lcm(x,j))
        sum[i]=sum[i-1]+f[i];
    }
}

int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    init();
    int t,icase=0,n;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        printf("Case %d: %llu\n",++icase,sum[n]);
    }
    return 0;
}