[POI2007]ZAP-Queries

嘟嘟嘟


挺好的题

\[\begin{align*} ans &= \sum_{i = 1} ^ {a} \sum_{j = 1} ^ {b} [gcd(i, j) = d] \\ &= \sum_{i = 1} ^ {\lfloor \frac{a}{d} \rfloor} \sum_{j = 1} ^ {\lfloor \frac{b}{d} \rfloor} [gcd(i, j) = 1] \\ \end{align*}\]

令\(n = \lfloor \frac{a}{d} \rfloor\)，\(m = \lfloor \frac{b}{d} \rfloor\)，根据莫比乌斯函数：\(\sum_{d | n} \mu(d) = [n = 1]\)，接着化简

\[\begin{align*} ans &= \sum_{i = 1} ^ {n} \sum_{j = 1} ^ {m} \sum_{d' | gcd(i, j)} \mu(d') \\ &= \sum_{d'} \sum_{i = 1, d' | i} ^ {n} \sum_{j = 1, d' | i} ^ {m} \mu(d') \\ &= \sum_{d'} \lfloor \frac{n}{d'} \rfloor \lfloor \frac{m}{d'} \rfloor \mu(d') \end{align*}\]

化简到这里，就可以用数论分块的思想，枚举\(d'\)的时候成块的求\(\mu(d')\)的值了。
所以要先预处理\(\mu(i)\)的前缀和。

#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<vector>
#include<stack>
#include<queue>
using namespace std;
#define enter puts("") 
#define space putchar(' ')
#define Mem(a, x) memset(a, x, sizeof(a))
#define rg register
typedef long long ll;
typedef double db;
const int INF = 0x3f3f3f3f;
const db eps = 1e-8;
const int maxn = 5e4 + 5;
inline ll read()
{
  ll ans = 0;
  char ch = getchar(), last = ' ';
  while(!isdigit(ch)) last = ch, ch = getchar();
  while(isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - '0', ch = getchar();
  if(last == '-') ans = -ans;
  return ans;
}
inline void write(ll x)
{
  if(x < 0) x = -x, putchar('-');
  if(x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}

int n;
int prime[maxn], v[maxn], phi[maxn], mu[maxn];
ll sum[maxn];
void init()
{
  phi[1] = mu[1] = 1;
  for(int i = 2; i < maxn; ++i)
    {
      if(!v[i]) v[i] = i, prime[++prime[0]] = i, phi[i] = i - 1, mu[i] = -1;
      for(int j = 1; i * prime[j] < maxn && j <= prime[0]; ++j)
	{
	  int k = i * prime[j];
	  v[k] = prime[j];
	  if(i % prime[j] == 0)
	    {
	      phi[k] = prime[j] * phi[i];
	      mu[k] = 0;
	      break;
	    }
	  else phi[k] = (prime[j] - 1) * phi[i], mu[k] = -mu[i];
	}
    }
  for(int i = 1; i < maxn; ++i) sum[i] = sum[i - 1] + mu[i];
}
ll solve(int n, int m)
{
  ll ret = 0; ll Min = min(n, m);
  for(int l = 1, r; l <= Min; l = r + 1)
    {
      r = min(n / (n / l), m / (m / l));
      ret += (sum[r] - sum[l - 1]) * (n / l) * (m / l);
    }
  return ret;
}

int main()
{
  init();
  int T = read();
  while(T--)
    {
      int n = read(), m = read(), d = read();
      write(solve(n / d, m / d)), enter;
    }
  return 0;
}

标签：lfloor,POI2007,rfloor,mu,maxn,Queries,include,sum,ZAP
来源： https://blog.51cto.com/u_15234622/2831207