1363 最小公倍数之和
推式子
【数论|51 NOD 1363 最小公倍数之和 (欧拉函数思维应用)】 ∑ i = 1 n l c m ( i , n ) = n ∑ i = 1 n i g c d ( i , n ) = n ∑ d ∣ n ∑ i = 1 n i d ( g c d ( i , n ) = = d ) = n ∑ d ∣ n ∑ i = 1 n d i ( g c d ( i , n d ) = = 1 ) = n ∑ d ∣ n d ? ( d ) + ( d = = 1 ) 2 \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 \phi(d) + (d == 1)} {2}\\ i=1∑n?lcm(i,n)=ni=1∑n?gcd(i,n)i?=nd∣n∑?i=1∑n?di?(gcd(i,n)==d)=nd∣n∑?i=1∑dn??i(gcd(i,dn?)==1)=nd∣n∑?2d?(d)+(d==1)?
接下来就是筛出 n \sqrt n n ?内的质数,再通过递归算出所有的因数,然后加上所有的因数对答案的贡献,具体细节在代码中描述。
代码
/*
Author : lifehappy
*/
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include #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 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为1的情况。
return ;
}
solve(pos + 1, n, phi);
//不选这个数的情况。
n *= fac[pos], phi *= (fac[pos] - 1);
//第一次选这个数要单独考虑,当两个数互质的时候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];
//这里就是不互质的情况了
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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