给出一个n，求1-n这n个数，同n的最小公倍数的和。
例如：n = 6，1,2,3,4,5,6 同6的最小公倍数分别为6,6,6,12,30,6，加在一起 = 66。
由于结果很大，输出Mod 1000000007的结果
ai <= 1e9
公式推导如上。
最后只需要预处理出质因子，用积性函数的性质在线处理即可。
#pragma warning(disable:4996) #include<iostream> #include<algorithm> #include<bitset> #include<tuple> #include<unordered_map> #include<fstream> #include<iomanip> #include<string> #include<cmath> #include<cstring> #include<vector> #include<map> #include<set> #include<list> #include<queue> #include<stack> #include<sstream> #include<cstdio> #include<ctime> #include<cstdlib> #define INF 0x3f3f3f3f #define inf 0x7FFFFFFF #define MOD 1000000007 #define moD 1000000003 #define pii pair<int,string> #define eps 1e-8 #define equals(a,b) (fabs(a-b)<eps) #define bug puts("bug") #define re register #define fi first #define se second const int maxn = 1e6 + 5; const double Inf = 10000.0; const double PI = acos(-1.0); typedef long long ll; typedef unsigned long long ull; using namespace std; int prime[maxn]; int vis[maxn]; int cnt; int euler_sieve(int n) { for (int i = 2; i <= n; i++) { if (!vis[i]) prime[cnt++] = i; for (int j = 0; j < cnt; j++) { if (i * prime[j] > n) break; vis[i * prime[j]] = 1; if (i % prime[j] == 0) break; } } return cnt; } ll quickPower(ll a, ll b, ll m) { ll base = a; ll ans = 1; while (b) { if (b & 1) { ans *= base; ans %= m; } base *= base; base %= m; b >>= 1; } return ans; } ll inv2; ll solve(ll n) { ll ans = 1; for (int i = 0; prime[i] * prime[i] <= n && i <= cnt; i++) { if (n % prime[i] == 0) { ll now = 1ll, sum = 1ll; while (n % prime[i] == 0) { n /= prime[i]; now *= prime[i]; sum += (now - now / prime[i]) * now; } ans = ans * sum % MOD; } } if (n > 1) ans = ans * (1ll + n * (n - 1) % MOD) % MOD; return ans + 1ll; } int main() { euler_sieve(40000); int T; inv2 = quickPower(2ll, MOD - 2, MOD); scanf("%d", &T); while (T--) { ll n; scanf("%lld", &n); printf("%lld\n", solve(n) * (n * inv2 % MOD) % MOD); } }