bzoj3529: [Sdoi2014]数表

题目链接

bzoj3529: [Sdoi2014]数表

题解

令\(d(x)\)表示\(x\)的约数和

就是求这个\[\sum_{i = 1}^n\sum_{j = 1}^{m} d(gcd(i,j)) \leq a\]

首先，我们不考虑a

另\(f(x)=\sum_i^n\sum_j^m gcd(i,j) == x\)

那么$f(x) = \sum_{x|p}\mu(\frac{p}{x})\lfloor \frac{n}{p}\rfloor\lfloor\frac{m}{p}\rfloor $

则答案就是

\[\sum_x^{min(n,m)}f(x)d(x)\]

\[=\sum_x^{min(n,m)}d(x)\sum_{x|p} \mu(\frac{p}{x})\lfloor \frac{n}{p}\rfloor\lfloor\frac{m}{p}\rfloor\]

\[\sum_p^{min(n,m)}\lfloor \frac{n}{p}\rfloor\lfloor\frac{m}{p}\rfloor
\sum_{x|p}\mu(\frac{p}{x})d(x)\]

对于\(d()\)我们可以nlogn筛出来

考虑a的限制，我们可以离线做

对于\(\sum_{x|p}\mu(\frac{p}{x})d(x)\)这部分

对于询问a的排序，对于\(d(i)\)排序，对于每次询问，把\(f(i)\leq a\)的每个\(i\)按上式丢到一个bit里维护一下前缀和,就可以偷税的数论分块了

复杂度\(O(\sqrt {n} logn + nlog^2n)\)

代码

#include <queue>
#include <cstdio>
#include <algorithm>
inline int read() {
int x = 0;
char c = getchar();
while(c < '0' || c > '9')c = getchar();
while(c <= '9' && c >= '0')x = x * 10 + c - '0',c = getchar();
return x;
}
const int mod = (1 << 31);
const int maxn = 100007;
int p[maxn],mu[maxn],mx; bool vis[maxn];
struct node {
int n,m,a,id;
bool operator < (const node & k)const {
return a < k.a;
}
} q[maxn];
std::pair<int,int>F[maxn];
void getmu() {
int size = mx,num = 0;
mu[1] = 1;
for(int i = 2;i <= size;++ i) {
if(!vis[i]) p[++ num] = i,mu[i] = -1;
for(int j = 1;j <= num && i * p[j] <= size;++ j) {
vis[i * p[j]] = true;
if(i % p[j] == 0) {
mu[i * p[j]] = 0; break;
} mu[i * p[j]] = -mu[i];
}
}
for(int i = 1;i <= size;++ i)
for(int j = i;j <= size;j += i)
F[j].first += i;
for(int i = 1;i <= size;++ i)F[i].second = i;
}
int ans[maxn];
#define lowbit(x) x&(-x)
int bit[maxn << 1];
void add(int x,int num) {
for(int i = x;i <= mx; i += lowbit(i))bit[i] += num;
}
int query(int x) {
int ret = 0;
for(int i = x;i; i -= lowbit(i)) ret += bit[i];
return ret;
}
void solve(int x) {
int n = q[x].n,m = q[x].m;
for(int i = 1,j;i <= q[x].n;i = j + 1) {
j = std::min(m / (m / i),n / (n/i));
ans[q[x].id] += (n / i) * (m / i) * (query(j) - query(i - 1));
}
}
int main() {
int k = read();
for(int i = 1;i <= k;++ i) {
q[i].n = read(),q[i].m = read(),q[i].a = read();
if(q[i].n > q[i].m) std::swap(q[i].n,q[i].m);
q[i].id = i; mx = std::max(mx,q[i].n);
}
getmu();
std::sort(q + 1,q + k + 1);
std::sort(F + 1,F + mx + 1);
for(int i = 1,now = 0;i <= k;++ i) {
for(;now + 1 <= mx && F[now + 1].first <= q[i].a;) {
now ++;
for(int j = F[now].second;j <= mx;j += F[now].second) {
//printf("%d\n",now);
add(j,F[now].first * mu[j / F[now].second]);
}
}
solve(i);
}
//printf("%d\n",mod);
for(int i = 1;i <= k;++ i) {
printf("%d\n",ans[i] & 0x7fffffff );
}
return 0;
}