[ans=sum_{i=1}^nlcm(i,n)
]
[egin{aligned}
ans
& =sum_{i=1}^nlcm(i,n)
\ & =sum_{i=1}^nfrac{icdot n}{gcd(i,n)}
\ & =n imes sum_{i=1}^nfrac{i}{gcd(i,n)}
\ & =n imessum_{i=1}^nisum_{d=1}^ifrac{[gcd(i,n)=d]}{d}
\ & =n imes sum_{d|n}sum_{i=1}^{leftlfloordfrac{n}{d}
ight
floor}icdot[gcd(i,frac{n}{d})=1]
end{aligned}
]
我们看后面的东西
设
[g(n)=sum_{i=1}^nicdot [gcd(i,n)=1]
]
我们知道更相减损术
[gcd(a,b)=gcd(a,a-b)
]
所以
[gcd(i,d)=1Leftrightarrow gcd(d-i,d)=1
]
所以
对于每一个i，都有另一个d-i与其对应，而且两个数的和是个定值！
所以
[g(n)=frac{varphi(n)}{2} imes d
]
但是g(1)是个特例，所以要特别处理
g出来了之后我们再预处理答案，核心代码如下
for (int i = 1; i <= N; i++) {
for (int j = i; j <= N; j += i) {
S[j] += (g[i] * i + 1) >> 1;
}
}
这样就可以做到了。
#include<cstdio>
#define Starseven main
#define ll long long
namespace lyt {
void read(int &x){
char ch=getchar();int re=0,op=1;
while(ch<'0'||ch>'9'){if(ch=='-') op=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){re=(re<<3)+(re<<1)+ch-'0';ch=getchar();}
x = re * op;
return ;
}
void read(long long &x){
char ch=getchar();long long re=0,op=1;
while(ch<'0'||ch>'9'){if(ch=='-') op=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){re=(re<<3ll)+(re<<1ll)+ch-'0';ch=getchar();}
x = re * op;
return ;
}
void write(int x){
if(x<0){putchar('-');x=-x;}
if(x>9) write(x/10);
putchar(x%10+'0');
return ;
}//记得自己加空格和换行
void write(long long x){
if(x<0){putchar('-');x=-x;}
if(x>9) write(x/10);
putchar(x%10+'0');
return ;
}//记得自己加空格和换行
int max(int x,int y){return x<y?y:x;}
int min(int x,int y){return x<y?x:y;}
int abs(int x){return x<0?-x:x;}
long long max(long long x,long long y){return x<y?y:x;}
long long min(long long x,long long y){return x<y?x:y;}
long long abs(long long x){return x<0?-x:x;}
double abs(double x){return x<0?-x:x;}
void swap(int &a,int &b) {a ^= b ^= a ^= b;}
void swap(long long &a,long long &b) {a ^= b ^= a ^= b;}
}using namespace lyt;
const int N = 1e6;
int prime[N + 20], num;
ll S[N + 20], g[N + 20];
bool vis[N + 20];
void Init(void) {
g[1] = 1;
for (int i = 2; i <= N; i++) {
if(!vis[i]) {
g[i] = i - 1;
prime[++num] = i;
}
for (int j = 1; j <= num && prime[j] * i <= N; j++) {
int x = prime[j] * i;
vis[x] = true;
if(i % prime[j] == 0) {
g[x] = g[i] * prime[j];
break;
}
g[x] = g[i] * g[prime[j]];
}
}
return ;
}
int Starseven(void) {
int t;
read(t);
Init();
for (int i = 1; i <= N; i++) {
for (int j = i; j <= N; j += i) {
S[j] += (g[i] * i + 1) >> 1;//我这里没有特判1，而是用了一点小技巧
}
}
while(t--) {
int n;
read(n);
write(S[n] * n * 1ll);
puts("");
}
return 0;
}