[HAOI2011]Problem b

嘟嘟嘟


看到求区间的，一个很好的思路就是转换成前缀和相减。那么这道题就是二维前缀和。
容易列出

\[\sum _ {i = 1} ^ {n} \sum _{j = 1} ^ {m} [gcd(i, j) = k] \]

然后就是套路的推导了，跟这道题一模一样，看我的题解吧[POI2007]ZAP-Queries。

#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 In inline
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 a, b, c, d, k;

int prim[maxn], v[maxn], mu[maxn];
ll sum[maxn];
In void init()
{
  mu[1] = 1;
  for(int i = 2; i < maxn; ++i)
    {
      if(!v[i]) v[i] = i, prim[++prim[0]] = i, mu[i] = -1;
      for(int j = 1; j <= prim[0] && prim[j] * i < maxn; ++j)
	{
	  v[i * prim[j]] = prim[j];
	  if(i % prim[j] == 0)
	    {
	      mu[i * prim[j]] = 0;
	      break;
	    }
	  else mu[i * prim[j]] = -mu[i];
	}
    }
  for(int i = 1; i < maxn; ++i) sum[i] = sum[i - 1] + mu[i];
}
In ll solve(int n, int m)
{
  n /= k; m /= k;
  if(n > m) swap(n, m);
  ll ret = 0;
  for(int l = 1, r; l <= n; 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();
  n = read();
  for(int i = 1; i <= n; ++i)
    {
      a = read(), b = read(), c = read(), d = read(), k = read();
      write(solve(b, d) - solve(b, c - 1) - solve(d, a - 1) + solve(a - 1, c - 1)), enter;
    }
  return 0;
}