P6810 「MCOI-02」Convex Hull 凸包

思路

$$\begin{gathered} \sum _{i=1}^n\sum _{j=1}^m\tau (i)\tau (j)\tau (gcd(i,j)) \\ \sum _{d=1}^n\tau (d)\sum _{i=1}^{\frac{n}{d}}\sum _{j=1}^{\frac{m}{d}}\tau (id)\tau (id)[gcd(i,j)==1] \\ \sum _{d=1}^n\tau (d)\sum _{k=1}^{\frac{n}{d}}\mu (k)\sum _{i=1}^{\frac{n}{dk}}\sum _{j=1}^{\frac{m}{dk}}\tau (idk)\tau (idk) \\ t=kd \\ \sum _{t=1}^n\sum _{i=1}^{\frac{n}{t}}\tau (it)\sum _{j=1}^{\frac{m}{t}}\tau (jt)\sum _{d\mid t}\tau (d)\mu (\frac{t}{d}) \\ \sum _{d\mid t}\tau (d)\mu (\frac{t}{d})=\tau ?\mu ,有\tau (n)=\sum _{d\mid n}=\sum _{d\mid n}1\times 1=I?I \\ 所以有\tau ?\mu =I?I?\mu =I??=I=1 \\ 原式:\sum _{t=1}^n\sum _{i=1}^{\frac{n}{t}}\tau (it)\sum _{j=1}^{\frac{m}{t}}\tau (jt) \\ 所以只要預處理出后面的就可以直接O(n)求解了。 \end{gathered}$$

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; }#define uint unsigned int uint seed; inline uint getnext(){seed^=seed<<13;seed^=seed>>17;seed^=seed<<5;return seed; }const int N = 2e6 + 10;int prime[N], cnt, n, m, mod;ll tau[N], f1[N], f2[N], f[N];bool st[N];void init() {tau[1] = 1;for(int i = 2; i < N; i++) {tau[i] = 1;if(!st[i]) prime[cnt++] = i;for(int j = 0; j < cnt && 1ll * i * prime[j] < N; j++) {st[i * prime[j]] = 1;if(i % prime[j] == 0) break;}}for(int i = 0; i < cnt; i++) {for(int j = 1; 1ll * j * prime[i] < N; j++) {tau[j * prime[i]] = (tau[j * prime[i]] + tau[j]) % mod;}} }int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);n = read(), m = read(), mod = read();init();if(n > m) swap(n, m);for(int i = 1; i <= m; i++) {if(i <= n) f1[i] = tau[i];f2[i] = tau[i];}for(int i = 0; i < cnt; i++) {for(int j = m / prime[i]; j >= 1; j--) {f2[j] = (f2[j] + f2[j * prime[i]]) % mod;}}for(int i = 0; i < cnt; i++) {for(int j = n / prime[i]; j >= 1; j--) {f1[j] = (f1[j] + f1[j * prime[i]]) % mod;}}ll ans = 0;for(int i = 1; i <= n; i++) {ans = (ans + 1ll * f1[i] * f2[i] % mod) % mod;}printf("%lld\n", ans);return 0; }

總結

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