2226: [Spoj 5971] LCMSum
Time Limit: 20 Sec Memory Limit: 259 MB
Submit: 1949 Solved: 852
[Submit][Status][Discuss]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
[Submit][Status][Discuss]
一个比较有用的式子：$$f(n)=\sum_{i=1}^{n}[gcd(i,n)=1]i$$$$f(1)=1\quad f(n)=\frac{\phi(n)*n}{2}$$
然后按照套路化式子即可：https://blog.sengxian.com/solutions/bzoj-2226
线性筛WA两次，怎么回事啊？
1 #include<cstdio> 2 #include<algorithm> 3 #define rep(i,l,r) for (int i=l; i<=r; i++) 4 typedef long long ll; 5 using namespace std; 6 7 const int N=1000003; 8 int n,T,tot,p[N],phi[N]; 9 bool b[N]; 10 ll g[N]; 11 12 void pre(){ 13 phi[1]=1; 14 for (int i=2; i<N; i++){ 15 if (!b[i]) p[++tot]=i,phi[i]=i-1; 16 for (int j=1; j<=tot && i*p[j]<N; j++){ 17 int t=i*p[j]; b[t]=1; 18 if (i%p[j]) phi[t]=(p[j]-1)*phi[i]; 19 else { phi[t]=p[j]*phi[i]; break; } 20 } 21 } 22 for (int i=1; i<N; i++) for (int j=i; j<N; j+=i) g[j]+=1ll*phi[i]*i; 23 } 24 25 int main(){ 26 freopen("bzoj2226.in","r",stdin); 27 freopen("bzoj2226.out","w",stdout); 28 pre(); 29 for (scanf("%d",&T); T--; ) scanf("%d",&n),printf("%lld\n",(n==1)?1:((n==2)?4:(g[n]+1)*n/2)); 30 return 0; 31 }
