HAOI2011 Problem b

传送门

做过上一道题之后，这个题就没啥难度了。就是加了个枚举的下界。

就像维护二维前缀和一样，直接把结果加加减减即可，具体方法和上一题一样。直接看代码。

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
#include<cmath>
#include<set>
#include<vector>
#include<map>
#include<queue>
#define rep(i,a,n) for(int i = a;i <= n;i++)
#define per(i,n,a) for(int i = n;i >= a;i--)
#define enter putchar('\n')
#define fr friend inline
#define y1 poj
#define mp make_pair
#define pr pair<int,int>
#define fi first
#define sc second
#define pb push_back

using namespace std;
typedef long long ll;
const int M = 100005;
const int INF = 1000000009;
const double eps = 1e-7;

int read()
{
    int ans = 0,op = 1;char ch = getchar();
    while(ch < '0' || ch > '9') {if(ch == '-') op = -1;ch = getchar();}
    while(ch >= '0' && ch <= '9') ans = ans * 10 + ch - '0',ch = getchar();
    return ans * op;
}

int a,b,c,d,k,p[M],mu[M],tot,sum[M],n;
bool np[M];

void euler()
{
   np[1] = 1,mu[1] = 1;
   rep(i,2,M-2)
   {
      if(!np[i]) p[++tot] = i,mu[i] = -1;
      for(int j = 1;i * p[j] <= M-2;j++)
      {
     np[i * p[j]] = 1;
     if(!(i % p[j])) break;
     mu[i * p[j]] = -mu[i];
      }
   }
   rep(i,1,M-2) sum[i] = sum[i-1] + mu[i];
}

int solve(int a,int b)
{
   a /= k,b /= k;
   int m = min(a,b),ans = 0;
   for(int i = 1,j;i <= m;i = j + 1)
   {
      j = min(a / (a / i),b / (b / i));
      ans += (a / i) * (b / i) * (sum[j] - sum[i-1]);
   }
   return ans;
}

int main()
{
   euler();
   n = read();
   rep(i,1,n)
   {
      a = read(),b = read(),c = read(),d = read(),k = read();
      printf("%d\n",solve(b,d) + solve(a-1,c-1) - solve(a-1,d) - solve(c-1,b));
   }
   return 0;
}