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数论|E. Number Challenge

数论codeforces

E. Number Challenge 推式子
∑ i = 1 a ∑ j = 1 b ∑ k = 1 c σ ( i j k ) = ∑ i = 1 a ∑ j = 1 b ∑ k = 1 c ∑ x ∣ i ∑ y ∣ j ∑ z ∣ k ( g c d ( x , y ) = 1 ) ( g c d ( x , z ) = 1 ) ( g c d ( y , z ) = 1 ) = ∑ x = 1 a ∑ y = 1 b ∑ z = 1 c ? a x ? ? b y ? ? c z ? ( g c d ( x , y ) = 1 ) ( g c d ( x , z ) = 1 ) ( g c d ( y , z ) = 1 ) = ∑ d = 1 a μ ( d ) ∑ x = 1 ? a d ? ? a d x ? ∑ y = 1 ? b d ? ? b d y ? ∑ z = 1 c ? c d ? ( g c d ( x , z ) = 1 ) ( g c d ( y , z ) = 1 ) = ∑ d = 1 a μ ( d ) ∑ z = 1 c ? c d ? ∑ x = 1 ? a d ? ? a d x ? ( g c d ( x , z ) = = 1 ) ∑ y = 1 ? b d ? ? b d x ? ( g c d ( y , z ) = = 1 ) \sum_{i = 1} ^{a} \sum_{j = 1} ^{b} \sum_{k = 1} ^{c} \sigma(ijk)\\ = \sum_{i = 1} ^{a} \sum_{j = 1} ^{b} \sum_{k = 1} ^{c} \sum_{x \mid i} \sum_{y \mid j} \sum_{z \mid k}(gcd(x,y) = 1)(gcd(x, z) = 1)(gcd(y, z) = 1)\\ = \sum_{x = 1} ^{a} \sum_{y = 1} ^{b} \sum_{z = 1} ^{c} \lfloor\frac{a}{x}\rfloor \lfloor \frac{b}{y}\rfloor \lfloor \frac{c}{z}\rfloor(gcd(x, y) = 1)(gcd(x, z) = 1)(gcd(y, z) = 1)\\ = \sum_{d = 1} ^{a} \mu(d) \sum_{x = 1} ^{\lfloor \frac{a}{d} \rfloor} \lfloor\frac{a}{dx}\rfloor \sum_{y = 1} ^{\lfloor \frac{b}{d}\rfloor} \lfloor \frac{b}{dy} \rfloor \sum_{z = 1} ^{c} \lfloor\frac{c}{d} \rfloor (gcd(x, z) = 1)(gcd(y, z) = 1)\\ = \sum_{d = 1} ^{a} \mu(d) \sum_{z = 1} ^{c} \lfloor \frac{c}{d} \rfloor \sum_{x = 1} ^{\lfloor \frac{a}{d}\rfloor} \lfloor \frac{a}{dx} \rfloor (gcd(x, z) == 1) \sum_{y = 1} ^{\lfloor \frac{b}{d}\rfloor}\lfloor\frac{b}{dx}\rfloor(gcd(y, z) == 1)\\ i=1∑a?j=1∑b?k=1∑c?σ(ijk)=i=1∑a?j=1∑b?k=1∑c?x∣i∑?y∣j∑?z∣k∑?(gcd(x,y)=1)(gcd(x,z)=1)(gcd(y,z)=1)=x=1∑a?y=1∑b?z=1∑c??xa???yb???zc??(gcd(x,y)=1)(gcd(x,z)=1)(gcd(y,z)=1)=d=1∑a?μ(d)x=1∑?da????dxa??y=1∑?db????dyb??z=1∑c??dc??(gcd(x,z)=1)(gcd(y,z)=1)=d=1∑a?μ(d)z=1∑c??dc??x=1∑?da????dxa??(gcd(x,z)==1)y=1∑?db????dxb??(gcd(y,z)==1)
【数论|E. Number Challenge】然后就是的 n 2 log ? n n ^ 2 \log n n2logn乱搞了。
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 = 2e3 + 10, mod = 1073741824; int g[N][N], prime[N], cnt; bool st[N]; ll mu[N], a, b, c; 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]; } } }int gcd(int a, int b) { if(!b) return a; if(g[a][b]) return g[a][b]; return g[a][b] = gcd(b, a % b); }void get_gcd() { for(int i = 1; i <= c; i++) { for(int j = 1; j <= c; j++) { g[i][j] = gcd(i, j); } } }void Sort(ll & a, ll & b, ll & c) { ll A = a, B = b, C = c; a = min({A, B, C}); c = max({A, B, C}); b = A + B + C - a - c; }ll f(ll n, ll m) { ll ans = 0; for(int i = 1; i <= n; i++) { if(g[i][m] == 1) { ans += n / i; } } ans = (ans + mod) % mod; return ans; }int main() { // freopen("in.txt", "r", stdin); // freopen("out.txt", "w", stdout); // ios::sync_with_stdio(false), cin.tie(0), cout.tie(0); a = read(), b = read(), c = read(); Sort(a, b, c); init(); get_gcd(); ll ans = 0; for(int i = 1; i <= a; i++) { for(int j = 1; j <= b; j++) { if(g[i][j] == 1) { ans = (ans + mu[j] * (a / i) % mod * f(b / j, i) % mod * f(c / j, i) % mod + mod) % mod; } } } cout << ans << endl; return 0; }

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