HDU 4279 Number（找规律）

Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4288    Accepted Submission(s): 1066

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

21 11 10

 

Sample Output

04
Hint
 For the second case, the real numbers are 6,8,9,10. 
 

 

Source

2012 ACM/ICPC Asia Regional Tianjin Online

 

题意： 给出一个f(x),表示不大于x的正整数里，不整除x且跟x有大于1的公约数的数的个数。定义F(x),为不大于x的正整数里，满足f(x)的值为奇数的数的个数。题目就是求这个F(x)

题解：打个表找规律，F(x)=x/2-1+(sqrt(x)%2? 0:-1) 推荐证明：http://blog.csdn.net/acdreamers/article/details/9730423  

注：sqrt（）存在精度问题，精度处理不好会wa.

#include<cstdio>#include<algorithm>#include<iostream>#include<cstring>#include<cmath>#define ll long longusing namespace std;ll Sqrt(ll l,ll r,ll a) {    ll mid=(l+r)/2;    if(l>r)        return r;    if(a/mid>mid)        return Sqrt(mid+1,r,a);    else if(a/mid<mid)        return Sqrt(l,mid-1,a);    else        return mid;}ll solve(ll r) {    if(r<6)return 0;    return r/2-2+(ll)Sqrt(1,r,r)%2;;}int main() {    int t;    cin>>t;    while(t--) {        ll l,r;        scanf("%I64d%I64d",&l,&r);        printf("%I64d\n",solve(r)-solve(l-1));    }    return 0;}