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

2
1 1
1 10

 

Sample Output

0
4

Hint

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

 

Source

field=problem&key=2012+ACM%2FICPC+Asia+Regional+Tianjin+Online&source=1&searchmode=source" style="color:rgb(26,92,200); text-decoration:none">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 long

using 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;
}