E. Number Challenge

推式子

$$\begin{gathered} \sum _{i=1}^a\sum _{j=1}^b\sum _{k=1}^c\sigma (ijk) \\ =\sum _{i=1}^a\sum _{j=1}^b\sum _{k=1}^c\sum _{x\mid i}\sum _{y\mid j}\sum _{z\mid k}(gcd(x,y)=1)(gcd(x,z)=1)(gcd(y,z)=1) \\ =\sum _{x=1}^a\sum _{y=1}^b\sum _{z=1}^c?\frac{a}{x}??\frac{b}{y}??\frac{c}{z}?(gcd(x,y)=1)(gcd(x,z)=1)(gcd(y,z)=1) \\ =\sum _{d=1}^a\mu (d)\sum _{x=1}^{?\frac{a}{d}?}?\frac{a}{dx}?\sum _{y=1}^{?\frac{b}{d}?}?\frac{b}{dy}?\sum _{z=1}^c?\frac{c}{d}?(gcd(x,z)=1)(gcd(y,z)=1) \\ =\sum _{d=1}^a\mu (d)\sum _{z=1}^c?\frac{c}{d}?\sum _{x=1}^{?\frac{a}{d}?}?\frac{a}{dx}?(gcd(x,z)==1)\sum _{y=1}^{?\frac{b}{d}?}?\frac{b}{dx}?(gcd(y,z)==1) \end{gathered}$$

然后就是的$n^2\log n$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; }

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