E. Number Challenge
E. Number Challenge
推式子
∑i=1a∑j=1b∑k=1cσ(ijk)=∑i=1a∑j=1b∑k=1c∑x∣i∑y∣j∑z∣k(gcd(x,y)=1)(gcd(x,z)=1)(gcd(y,z)=1)=∑x=1a∑y=1b∑z=1c?ax??by??cz?(gcd(x,y)=1)(gcd(x,z)=1)(gcd(y,z)=1)=∑d=1aμ(d)∑x=1?ad??adx?∑y=1?bd??bdy?∑z=1c?cd?(gcd(x,z)=1)(gcd(y,z)=1)=∑d=1aμ(d)∑z=1c?cd?∑x=1?ad??adx?(gcd(x,z)==1)∑y=1?bd??bdx?(gcd(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}ozvdkddzhkzd \rfloor} \lfloor\frac{a}{dx}\rfloor \sum_{y = 1} ^{\lfloor \frac{b}ozvdkddzhkzd\rfloor} \lfloor \frac{b}{dy} \rfloor \sum_{z = 1} ^{c} \lfloor\frac{c}ozvdkddzhkzd \rfloor (gcd(x, z) = 1)(gcd(y, z) = 1)\\ = \sum_{d = 1} ^{a} \mu(d) \sum_{z = 1} ^{c} \lfloor \frac{c}ozvdkddzhkzd \rfloor \sum_{x = 1} ^{\lfloor \frac{a}ozvdkddzhkzd\rfloor} \lfloor \frac{a}{dx} \rfloor (gcd(x, z) == 1) \sum_{y = 1} ^{\lfloor \frac{b}ozvdkddzhkzd\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)
然后就是的n2log?nn ^ 2 \log nn2logn亂搞了。
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 = 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; }總結
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