P3911 最小公倍数之和

推式子

$$\begin{gathered} \sum _{i=1}^n\sum _{j=1}^{n}lcm(a_i,a_j) \\ 下面的n=max(a_i)，c_i为i在原数组中出现的次数 \\ \sum _{i=1}^n\sum _{j=1}^n\frac{ij}{gcd(ij)}{c}_i{c}_j \\ =\sum _{d=1}^n\frac{1}{d}\sum _{i=1}^n\sum _{j=1}^{n}ij{c}_i{c}_j(gcd(i,j)==d) \\ =\sum _{d=1}^{n}d\sum _{i=1}^{\frac{n}{d}}\sum _{j=1}^{\frac{n}{d}}ij{c}_{id}{c}_{jd}\sum _{k\mid gcd(i,j)}\mu (k) \\ =\sum _{d=1}^{n}d\sum _{k=1}^{\frac{n}{d}}\mu (k)k^2\sum _{i=1}^{\frac{n}{kd}}\sum _{j=1}^{\frac{n}{kd}}ij{c}_{ikd}{c}_{jkd} \\ =\sum _{t=1}^{n}t{\left(\sum _{i=1}^{\frac{n}{t}}i{c}_{it}\right)}^2\sum _{d\mid t}d\mu (d) \end{gathered}$$

代码

/*Author : lifehappy
*/
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include <bits/stdc++.h>#define mp make_pair
#define pb push_back
#define endl '\n'
#define mid (l + r >> 1)
#define lson rt << 1, l, mid
#define rson rt << 1 | 1, mid + 1, r
#define ls rt << 1
#define rs rt << 1 | 1using namespace std;typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;const double pi = acos(-1.0);
const double eps = 1e-7;
const int inf = 0x3f3f3f3f;inline ll read() {ll f = 1, x = 0;char c = getchar();while(c < '0' || c > '9') {if(c == '-')    f = -1;c = getchar();}while(c >= '0' && c <= '9') {x = (x << 1) + (x << 3) + (c ^ 48);c = getchar();}return f * x;
}const int N = 5e4 + 10;ll sum[N], c[N], n, m;bool st[N];int prime[N], mu[N], cnt;void init() {mu[1] = 1;for(int i = 2; i < N; i++) {if(!st[i]) {prime[cnt++] = i;mu[i] = -1;}for(int j = 0; j < cnt && i * prime[j] < N; j++) {st[i * prime[j]] = 1;if(i % prime[j] == 0) break;mu[i * prime[j]] = -mu[i];}}for(int i = 1; i < N; i++) {for(int j = i; j < N; j += i) {sum[j] += i * mu[i];}}
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);init();n = read(), m = 0;for(int i = 1; i <= n; i++) {ll x = read();m = max(x, m);c[x]++;}ll ans = 0;for(ll t = 1; t <= m; t++) {ll res = 0;for(ll i = 1; i <= m / t; i++) {res += 1ll * i * c[i * t];}ans += t * res * res * sum[t];}printf("%lld\n", ans);return 0;
}

C - LCMs

推式子

$$\begin{gathered} \sum _{i=1}^{n-1}\sum _{j=i+1}^{n}lcm(a_i,a_j) \\ =\frac{{\sum }_{i=1}^n{\sum }_{j=1}^{n}lcm(a_i,a_j)-{\sum }_{i=1}^n{a}_i}{2} \\ 求\sum _{i=1}^n\sum _{j=1}^{n}lcm(a_i,a_j) \\ 下面的n=max(a_i)，c_i为i在原数组中出现的次数 \\ \sum _{i=1}^n\sum _{j=1}^n\frac{ij}{gcd(ij)}{c}_i{c}_j \\ =\sum _{d=1}^n\frac{1}{d}\sum _{i=1}^n\sum _{j=1}^{n}ij{c}_i{c}_j(gcd(i,j)==d) \\ =\sum _{d=1}^{n}d\sum _{i=1}^{\frac{n}{d}}\sum _{j=1}^{\frac{n}{d}}ij{c}_{id}{c}_{jd}\sum _{k\mid gcd(i,j)}\mu (k) \\ =\sum _{d=1}^{n}d\sum _{k=1}^{\frac{n}{d}}\mu (k)k^2\sum _{i=1}^{\frac{n}{kd}}\sum _{j=1}^{\frac{n}{kd}}ij{c}_{ikd}{c}_{jkd} \\ =\sum _{t=1}^{n}t{\left(\sum _{i=1}^{\frac{n}{t}}i{c}_{it}\right)}^2\sum _{d\mid t}d\mu (d) \end{gathered}$$

代码

/*Author : lifehappy
*/
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include <bits/stdc++.h>#define mp make_pair
#define pb push_back
#define endl '\n'
#define mid (l + r >> 1)
#define lson rt << 1, l, mid
#define rson rt << 1 | 1, mid + 1, r
#define ls rt << 1
#define rs rt << 1 | 1using namespace std;typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;const double pi = acos(-1.0);
const double eps = 1e-7;
const int inf = 0x3f3f3f3f;inline ll read() {ll f = 1, x = 0;char c = getchar();while(c < '0' || c > '9') {if(c == '-')    f = -1;c = getchar();}while(c >= '0' && c <= '9') {x = (x << 1) + (x << 3) + (c ^ 48);c = getchar();}return f * x;
}const int N = 1e6 + 10, mod = 998244353;ll sum[N], c[N], n, m, all;bool st[N];int prime[N], mu[N], cnt;ll quick_pow(ll a, ll n, ll mod) {ll ans = 1;while(n) {if(n & 1) ans = ans * a % mod;n >>= 1;a = a * a % mod;}return ans;
}void init() {mu[1] = 1;for(int i = 2; i < N; i++) {if(!st[i]) {prime[cnt++] = i;mu[i] = -1;}for(int j = 0; j < cnt && i * prime[j] < N; j++) {st[i * prime[j]] = 1;if(i % prime[j] == 0) break;mu[i * prime[j]] = -mu[i];}}for(int i = 1; i < N; i++) {for(int j = i; j < N; j += i) {sum[j] = (sum[j] + i * mu[i] % mod + mod) % mod;}}
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);init();n = read(), m = 0;for(int i = 1; i <= n; i++) {ll x = read();all = (all + x) % mod;m = max(x, m);c[x]++;}ll ans = 0;for(ll t = 1; t <= m; t++) {ll res = 0;for(ll i = 1; i <= m / t; i++) {res = (res + 1ll * i * c[i * t] % mod) % mod;}ans = (ans + t * res % mod * res % mod * sum[t] % mod) % mod;}printf("%lld\n", ((ans - all) * quick_pow(2, mod - 2, mod) % mod + mod) % mod);return 0;
}