转载请注明出处：http://blog.csdn.net/u012860063?
viewmode=contents
题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4279
|
|
NumberTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2936 Accepted Submission(s): 805
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
Sample Output
Source
Recommend
liuyiding
|
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)。
代码例如以下：（用C++提交WA了无数次，用G++一遍过）
#include<cstdio>
#include<cmath>
#include<iostream>
#include<algorithm>
using namespace std;
//大于4。并且不是偶数的平方数的偶数是real number
//奇数的平方的奇数是real number
__int64 calc(__int64 n)//计算小于等于n的real number的个数
{
if(n<=4)
return 0;
__int64 t=sqrt(n*1.0);
__int64 ans=(n-4)/2;//大于4的偶数的个数
if(t%2==0)
return ans;
else
return ans+1;
}
int main()
{
int T;
__int64 A,B;
scanf("%d",&T);
while(T--)
{
scanf("%I64d%I64d",&A,&B);
printf("%I64d
",calc(B)-calc(A-1));
}
return 0;
}版权声明：本文博客原创文章，博客，未经同意，不得转载。