题解 P1891 【疯狂 LCM】
\[ans=\sum_{i=1}^nlcm(i,n)
\]
\[\begin{aligned}
ans
& =\sum_{i=1}^nlcm(i,n)
\\ & =\sum_{i=1}^n\frac{i\cdot n}{gcd(i,n)}
\\ & =n\times \sum_{i=1}^n\frac{i}{gcd(i,n)}
\\ & =n\times\sum_{i=1}^ni\sum_{d=1}^i\frac{[gcd(i,n)=d]}{d}
\\ & =n\times \sum_{d|n}\sum_{i=1}^{\left\lfloor\dfrac{n}{d}\right\rfloor}i\cdot[gcd(i,\frac{n}{d})=1]
\end{aligned}
\]
我们看后面的东西
设
\[g(n)=\sum_{i=1}^ni\cdot [gcd(i,n)=1]
\]
我们知道更相减损术
\[gcd(a,b)=gcd(a,a-b)
\]
所以
\[gcd(i,d)=1\Leftrightarrow gcd(d-i,d)=1
\]
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
\[g(n)=\frac{\varphi(n)}{2}\times d
\]
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;
}
