hdu 4279 Number (规律题 2012 ACM/ICPC Asia Regional Tianjin Online )

 Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 867    Accepted Submission(s): 280

Problem Description

　 　Here are two numbers A and B (0 < A <= B). If B cannot be divisible by A, and A and B are not co-prime numbers, we define A as a special number of B.
　　For each x, f(x) equals to the amount of x’s special numbers.
　　For example, f(6)=1, because 6 only have one special number which is 4. And f(12)=3, its special numbers are 8,9,10.
　　When f(x) is odd, we consider x as a real number.
　　Now given 2 integers x and y, your job is to calculate how many real numbers are between them.

 

Input

　 　In the first line there is an integer T (T <= 2000), indicates the number of test cases. Then T line follows, each line contains two integers x and y (1 <= x <= y <= 2^63-1) separated by a single space.

 

Output

　　Output the total number of real numbers.

 

Sample Input

2 1 1 1 10

 

Sample Output

0 4 
Hint
 For the second case, the real numbers are 6,8,9,10. 
 

 

Source

2012 ACM/ICPC Asia Regional Tianjin Online

 

Recommend

liuyiding

 题解：

规律，自己在纸上写写，就能找到规律 ，

很容易发现 所谓的 real number  就是大于4的偶数 且 不是 偶数的 平方的 数，或者是奇数的平方。

所以如果 k*k<=n<(k+1)*(k+1).假如k是偶数，那么就是 (n-4)/2 ,因为偶数的平方和奇数的平方个数相等。

假如k是奇数，那么就是  (n-4)/2+1了，因为奇数的平方多一个。

 

其余就简单了。注意数据范围，用 long long

 1 #include<cstdio>
 2  #include<cstring>
 3  #include<cmath>
 4  #include<iostream>
 5  #include<algorithm>
 6  #include<set>
 7  #include<map>
 8  #include<queue>
 9  #include<vector>
10  #include<string>
11  #define Min(a,b) a<b?a:b
12  #define Max(a,b) a>b?a:b
13  #define CL(a,num) memset(a,num,sizeof(a));
14  #define eps  1e-12
15  #define inf 100000000
16  #define mx  10
17 
18  const double pi  = acos(-1.0);
19  const int  maxn = 105;
20  typedef   __int64  ll;
21  using namespace std;
22  ll cal(ll n)
23  {
24      if(n <= 4) return  0;
25      ll tp = sqrt(n);
26      if(tp&1)
27      {
28          return (n - 4)/2 + 1;
29 
30      }
31      else return (n - 4)/2 ;
32  }
33  int main()
34  {
35      int n;
36      int t;
37      ll a,b;
38      scanf("%d",&t);
39      while(t--)
40      {
41          scanf("%I64d%I64d",&a,&b);
42 
43          printf("%I64d\n",cal(b) - cal(a - 1));
44      }
45  }