1. 程式人生 > >洛谷 P3327 [SDOI2015]約數個數和【莫比烏斯反演】

洛谷 P3327 [SDOI2015]約數個數和【莫比烏斯反演】

前置技能:

d(ij)=x|iy|j[gcd(x,y)=1]
然後按照套路一頓亂推……好吧我是沒推出來= =
最終結果:
Ans=d=1min(n,m)μ(d)(x=1ndndx)(y=1mdmdy)
#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<string>
#include<map> #include<queue> #include<vector> using namespace std; #define ll long long int #define INF 0x3f3f3f3f const int maxn = 5e4 + 10; int n, m; bool vis[maxn]; int sum[maxn]; int prim[maxn]; int mu[maxn]; ll g[maxn]; int cnt; void get_mu(int n) { mu[1] = 1; for (int
i = 2; i <= n; i++) { if (!vis[i]) { mu[i] = -1; prim[++cnt] = i; } for (int j = 1; j <= cnt && prim[j] * i <= n; j++) { vis[i*prim[j]] = 1; if (i%prim[j] == 0) break; else mu[prim[j] * i] = -mu[i]; } } /*for (int j = 1; j <= cnt; j++) for (int i = 1; i*prim[j] <= n; i++) g[i*prim[j]] += mu[i];*/
for (int i = 1; i <= n; i++) sum[i] = sum[i - 1] + mu[i]; for (int i = 1; i <= n; i++) { ll ans = 0; for (int l = 1, r; l <= i; l = r + 1) { r = (i / (i / l)); ans += 1ll * (r - l + 1) * 1ll * (i / l); } g[i] = ans; } } int main() { int T; scanf("%d", &T); get_mu(50000); while (T--) { scanf("%d %d", &n, &m); ll res, ans; res = min(n, m); ans = 0; for (ll l = 1, r; l <= res; l = r + 1) { r = min(n / (n / l), m / (m / l)); ans += (sum[r] - sum[l - 1]) * 1ll * g[n / l] * 1ll * g[m / l]; } printf("%lld\n", ans); } return 0; } //_CRT_SECURE_NO_WARNINGS