hdu4279 Number-------天津网络赛 打表找规律
Number
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 492 Accepted Submission(s): 159
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
比赛的时候真是把神经绷得太紧了。
打表代码
[cpp]
#include<iostream>
#include<cstdlib>
#include<stdio.h>
#include<math.h>
using namespace std;
#define ll __int64
ll a[110];
int main()
{
a[1]=0;
for(int i=2;i<=50;i++)
{
int count=0;
for(int j=2;j<i;j++)
{
if(i%j==0) continue;
bool flag=false;
for(int k=2;k<j;k++)
{
if(j%k==0&&i%k==0)
{
flag=true;break;
}
}
if(flag&&(i%j!=0)) {
// cout<<i<<" "<<j<<"*"<<endl;
count++;
}
}
if(count&1) a[i]=a[i-1]+1;
else a[i]=a[i-1];
cout<<i<<" "<<a[i]<<endl;
}
}
规律：long long xx = (long long)sqrt(x * 1.0); if(xx & 1) { return ((x - 4) >> 1) + 1; } else { return ((x - 4) >> 1); }
补充：软件开发 , C++ ,
