1363 最小公倍數(shù)之和

推式子

$$\begin{gathered} \sum _{i=1}^{n}lcm(i,n) \\ =n\sum _{i=1}^n\frac{i}{gcd(i,n)} \\ =n\sum _{d\mid n}\sum _{i=1}^n\frac{i}{d}(gcd(i,n)==d) \\ =n\sum _{d\mid n}\sum _{i=1}^{\frac{n}{d}}i(gcd(i,\frac{n}{d})==1) \\ =n\sum _{d\mid n}\frac{d?(d)+(d==1)}{2} \end{gathered}$$

接下來就是篩出$\sqrt{n}$內(nèi)的質(zhì)數(shù)，再通過遞歸算出所有的因數(shù)，然后加上所有的因數(shù)對答案的貢獻(xiàn)，具體細(xì)節(jié)在代碼中描述。

代碼

/*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 = 1e5 + 10, mod = 1e9 + 7;int prime[N], cnt;bool st[N];void init() {for(int i = 2; i < N; i++) {if(!st[i]) prime[cnt++] = i;for(int j = 0; j < cnt && i * prime[j] < N; j++) {st[i * prime[j]] = 1;if(i % prime[j] == 0) break;}} }int fac[50], num[50], tot;ll ans;void solve(int pos, int n, int phi) {if(pos == tot + 1) {ans = (ans + 1ll * n * (phi + (n == 1)) / 2 % mod) % mod;//特判n為1的情況。return ;}solve(pos + 1, n, phi);//不選這個數(shù)的情況。n *= fac[pos], phi *= (fac[pos] - 1);//第一次選這個數(shù)要單獨考慮，當(dāng)兩個數(shù)互質(zhì)的時候phi[i * prime] = phi[i] * (prime - 1)solve(pos + 1, n, phi);for(int i = 1; i < num[pos]; i++) {n *= fac[pos], phi *= fac[pos];//這里就是不互質(zhì)的情況了solve(pos + 1, n, phi);} }int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);init();int T = read();while(T--) {tot = 0;int n = read(), m = n;for(int i = 0; prime[i] * prime[i] <= n; i++) {if(n % prime[i] == 0) {fac[++tot] = prime[i], num[tot] = 0;while(n % prime[i] == 0) {n /= prime[i];num[tot]++;}}}if(n != 1) {fac[++tot] = n, num[tot] = 1;}ans = 0;solve(1, 1, 1);printf("%lld\n", ans * m % mod);}return 0; }

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