function
题意：典型的莫比乌斯反演就不用说了。。。
公式不会推，看的题解=_= 链接
1 #include <bits/stdc++.h> 2 using namespace std; 3 #define ll long long 4 const ll mod = 1e9+7; 5 const ll maxn = 1e6+10; 6 7 ll pri[maxn]; 8 ll mu[maxn]; 9 map<ll, ll> hs; 10 11 void init(){ 12 ll cnt = 0; 13 memset(pri, 0, sizeof(pri)); 14 mu[1] = 1; 15 for(ll i = 2; i < maxn; i++){ 16 if(!pri[i]){ 17 pri[cnt++] = i; 18 mu[i] = -1; 19 } 20 for(ll j = 0; j < cnt; j++) { 21 ll t = i * pri[j]; 22 if(t >= maxn) break; 23 pri[t] = 1; 24 if(i % pri[j] == 0) { 25 mu[t] = 0; 26 break; 27 } else { 28 mu[t] = -mu[i]; 29 } 30 } 31 mu[i] =(mu[i]+mu[i-1])%mod; 32 } 33 } 34 35 ll sum(ll x){ 36 ll mod1 = 6LL*mod; 37 ll res = x*(x+1)%mod1*(2*x+1)%mod1/6; 38 ll mod2 = 2LL*mod; 39 ll temp = 3*x*(x+1)%mod2/2; 40 res = (res - temp + 2LL*x + mod) % mod; 41 return res; 42 } 43 44 ll cal(ll n){ 45 if(n < maxn) return mu[n]; 46 if(hs[n]) return hs[n]; 47 ll L = 2, R = 2; 48 ll ans = 1; 49 while(L <= n){ 50 ans = (ans - (cal(n/R)*(R-L+1)+mod)%mod + mod) %mod; 51 if(R == n) break; 52 L = R+1; 53 R = n/(n/L); 54 } 55 return hs[n] = ans; 56 } 57 58 ll solve(ll n){ 59 ll res = 0; 60 ll L = 1, R = 1; 61 while(L <= n){ 62 ll ans = cal(n/R) * ((sum(R)- sum(L-1) + mod)%mod) %mod; 63 res = ((ans+res)%mod + mod) %mod; 64 if(R == n) break; 65 L = R+1; 66 R = n/(n/L); 67 } 68 res = (res+mod)%mod; 69 return res; 70 } 71 72 int main(){ 73 init(); 74 int t; 75 scanf("%d", &t); 76 while(t--){ 77 ll n; 78 scanf("%lld", &n); 79 ll res = solve(n); 80 printf("%lld ", res); 81 } 82 return 0; 83 }
1 #include <bits/stdc++.h> 2 using namespace std; 3 #define ll long long 4 const ll mod = 1e9+7; 5 const ll maxn = 1e6+10; 6 7 ll pri[maxn]; 8 ll mu[maxn]; 9 10 const int H_SZ = 13131; 11 12 struct HS{ 13 ll val[H_SZ], res[H_SZ]; 14 int head[maxn], nex[maxn]; 15 int cnt; 16 void init(){ 17 memset(head, -1, sizeof(head)); 18 cnt = 0; 19 } 20 21 void add(ll x, ll y){ 22 int u = x%H_SZ; 23 val[cnt] = x; 24 res[cnt] = y; 25 nex[cnt] = head[u]; 26 head[u] = cnt++; 27 } 28 ll query(ll x){ 29 int u = x%H_SZ; 30 for(int i = head[u]; ~i; i = nex[i]){ 31 if(val[i] == x) return res[i]; 32 } 33 return H_SZ; 34 } 35 }hs; 36 37 void init(){ 38 ll cnt = 0; 39 memset(pri, 0, sizeof(pri)); 40 mu[1] = 1; 41 for(ll i = 2; i < maxn; i++){ 42 if(!pri[i]){ 43 pri[cnt++] = i; 44 mu[i] = -1; 45 } 46 for(ll j = 0; j < cnt; j++) { 47 ll t = i * pri[j]; 48 if(t >= maxn) break; 49 pri[t] = 1; 50 if(i % pri[j] == 0) { 51 mu[t] = 0; 52 break; 53 } else { 54 mu[t] = -mu[i]; 55 } 56 } 57 mu[i] =(mu[i]+mu[i-1])%mod; 58 } 59 } 60 61 inline ll sum(ll x){ 62 ll mod1 = 6LL*mod; 63 ll res = x*(x+1)%mod1*(2*x+1)%mod1/6; 64 ll mod2 = 2LL*mod; 65 ll temp = 3*x*(x+1)%mod2/2; 66 res = (res - temp + 2LL*x + mod) % mod; 67 return res; 68 } 69 70 ll cal(ll n){ 71 if(n < maxn) return mu[n]; 72 if(hs.query(n) != H_SZ) return hs.query(n); 73 ll L = 2, R = 2; 74 ll ans = 1; 75 while(L <= n){ 76 ans = (ans - (cal(n/R)*(R-L+1)+mod)%mod + mod) %mod; 77 if(R == n) break; 78 L = R+1; 79 R = n/(n/L); 80 } 81 hs.add(n, ans); 82 return ans; 83 } 84 85 ll solve(ll n){ 86 ll res = 0; 87 ll L = 1, R = 1; 88 while(L <= n){ 89 ll ans = sum(n/R) * ((cal(R)- cal(L-1) + mod)%mod) %mod; 90 res = ((ans+res)%mod + mod) %mod; 91 if(R == n) break; 92 L = R+1; 93 R = n/(n/L); 94 } 95 res = (res+mod)%mod; 96 return res; 97 } 98 99 int main(){ 100 init(); 101 hs.init(); 102 int t; 103 scanf("%d", &t); 104 while(t--){ 105 ll n; 106 scanf("%lld", &n); 107 ll res = solve(n); 108 printf("%lld ", res); 109 } 110 return 0; 111 }