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sigma))，它是个积性函数(筛法)，(n<m)

则题目要求的就是(sum_{i=1}^{n}sum_{j=1}^{m}f(gcd(i, j)))

考虑每个(gcd)的贡献，(sum_{i=1}^{n}f(i)sum_{j=1}^{lfloorfrac{n}{i} floor}mu(j)lfloorfrac{n}{i*j} floorlfloorfrac{m}{i*j} floor)

替换(i*j)就是(sum_{k=1}^{n}lfloorfrac{n}{k} floorlfloorfrac{m}{k} floorsum_{d|k}f(d)mu(frac{k}{d}))

但是我们有限制，就不能直接筛(sum_{d|k}f(d)mu(frac{k}{d}))

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(sum_{d|n}f(d)mu(frac{n}{d}))加进来，树状数组维护前缀和即可

只要预处理处(mu)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
", ans[i] < 0 ? ans[i] + INF + 1 : ans[i]);
    return 0;
}