Bzoj3529: [Sdoi2014]数表

题面

传送门

Sol

先不管a的限制

设f(n)表示f的约数和(据说是σ)，它是个积性函数(筛法)，n<m

则题目要求的就是∑ni=1∑mj=1f(gcd(i,j))

考虑每个gcd的贡献，∑ni=1f(i)∑⌊ni⌋j=1μ(j)⌊ni∗j⌋⌊mi∗j⌋

替换i∗j就是∑nk=1⌊nk⌋⌊mk⌋∑d|kf(d)μ(kd)

但是我们有限制，就不能直接筛∑d|kf(d)μ(kd)

所以考虑离线处理，把询问按a排序，每次把小于等于a的f同它的所有倍数n的∑d|nf(d)μ(nd)加进来，树状数组维护前缀和即可

只要预处理处μ和f就好了，见上面的筛法链接

# include <bits/stdc++.h>
# define RG register
# define IL inline
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;
const int _(1e5 + 1), INF(2147483647);

IL ll Read(){
RG ll x = 0, z = 1; RG char c = getchar();
for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
return x * z;
}

int prime[_], num, N, Q, id[_], ans[_], f[_], mu[_], sumd[_], powd[_], bit[_];
bool isprime[_];
struct Qry{
int n, m, a, id;
IL bool operator <(RG Qry B) const{  return a < B.a;  }
} qry[_];

IL void Prepare(){
isprime[1] = 1; id[1] = f[1] = mu[1] = 1;
for(RG int i = 2; i < N; ++i){
id[i] = i;
if(!isprime[i]){
prime[++num] = i; f[i] = i + 1; mu[i] = -1;
sumd[i] = 1 + i; powd[i] = i;
}
for(RG int j = 1; j <= num && i * prime[j] < N; ++j){
isprime[i * prime[j]] = 1;
if(i % prime[j]){
sumd[i * prime[j]] = 1 + prime[j]; powd[i * prime[j]] = prime[j];
f[i * prime[j]] = f[i] * f[prime[j]];
mu[i * prime[j]] = -mu[i];
}
else{
mu[i * prime[j]] = 0;
powd[i * prime[j]] = powd[i] * prime[j];
sumd[i * prime[j]] = sumd[i] + powd[i * prime[j]];
f[i * prime[j]] = f[i] / sumd[i] * sumd[i * prime[j]];
break;
}
}
}
}

IL bool Cmp(RG int x, RG int y){  return f[x] < f[y];  }

IL void Add(RG int x, RG int d){  for(; x < N; x += x & -x) bit[x] += d;  }

IL int Query(RG int x){  RG int ret = 0; for(; x; x -= x & -x) ret += bit[x]; return ret;  }

IL int Calc(RG int n, RG int m){
RG int ret = 0, lst = 0, now;
for(RG int i = 1, j; i <= n; i = j + 1){
j = min(n / (n / i), m / (m / i));
now = Query(j);
ret += (n / i) * (m / i) * (now - lst);
lst = now;
}
return ret;
}

int main(RG int argc, RG char* argv[]){
Q = Read();
for(RG int i = 1, n, m, a; i <= Q; ++i){
n = Read(); m = Read(); a = Read();
if(n > m) swap(n, m);
qry[i] = (Qry){n, m, a, i};
N = max(N, n + 1);
}
Prepare();
sort(qry + 1, qry + Q + 1); sort(id + 1, id + N, Cmp);
for(RG int i = 1, j = 1; i <= Q; ++i){
for(; j < N && f[id[j]] <= qry[i].a; ++j)
for(RG int k = id[j]; k < N; k += id[j])
Add(k, f[id[j]] * mu[k / id[j]]);
ans[qry[i].id] = Calc(qry[i].n, qry[i].m);
}
for(RG int i = 1; i <= Q; ++i) printf("%d\n", ans[i] < 0 ? ans[i] + INF + 1 : ans[i]);
return 0;
}