洛谷 P3327 [SDOI2015]約數個數和【莫比烏斯反演】
阿新 • • 發佈:2018-12-24
前置技能:
然後按照套路一頓亂推……好吧我是沒推出來= =
最終結果:
#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