HDU 4279 2012网络赛Number(数论 欧拉函数结论约数个数)

Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2500    Accepted Submission(s): 692

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

 

题目大意：题目先给你一个定义f(x)表示的是比<=x且不是x的约数并且与x互素的个数，如果f(x)为奇数，那么x可以算做特殊数，问你a~b之间有多少个特殊数。

  解题思路：除了1之外，没有与x互素且是x的约数的数字。所以就好办了。f(x)=x-约数个数-互素个数+1.  由欧拉函数值得到，phi(x)为偶数(x>2)。而一个数的约数的个数是由它素数分解幂数决定的，比如x=e1^p1*e2^p2.....x的约数个数为(p1+1)*(p2+1)*(...)...那么如果x的约数个数为奇数，p1,p2,..必须都为偶数，那么x必须为平方数。

下面开始讨论f(x)为奇数的情况:

1.x为奇数，约数个数需要为奇数，那么x为平方数。

2.x为偶数，约数个数需要为偶数，那么x不为平方数。

特殊的:f(1)=0,f(2)=0;

下面寻找1~x满足条件的情况:

a.偶数个数为x/2-1,减去1是因为除去了偶数2

b.偶数平方个数为sqrt(x)/2

c.是奇数,又是平方数个数为sqrt(x)-sqrt(x/2)-1   //平方数-偶数的平方数-奇数1

答案是a-b+c,得到x/2-2+sqrt(x)-sqrt(x)/2-sqrt(x)/2;

可以sqrt(x)分奇偶讨论:

res=x/2-2+sqrt(x)%2?1:0;

这个好像只有G++能A掉，用c++WA了无数次。。。

  题目地址：Number

AC代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<cmath>
using namespace std;

__int64 cal(__int64 x)
{
    __int64 ans=0;
    if(x<=5) return ans;
    __int64 p=(__int64)sqrt(x*1.0+0.5);
    ans=(x>>1)-2;
    if(p&1) ans++;
    return ans;
}

int main()
{
    int tes;
    scanf("%d",&tes);
    __int64 a,b;
    while(tes--)
    {
        scanf("%I64d%I64d",&a,&b);
        printf("%I64d\n",cal(b)-cal(a-1));
    }
    return 0;
}

开始是打表写的，下面是打表的代码：

int gcd(int m,int n)
{
    int t;
    while(n)
    {
        t=m%n;
        m=n;
        n=t;
    }
    return m;
}

int main()
{

    int i,j;
    int p[55];
    p[0]=0;
    for(i=1; i<=50; i++)
    {
        int cnt=0;
        for(j=2; j<i; j++)
            if(gcd(i,j)>1&&(i%j!=0))
                cnt++;
        if(cnt&1) p[i]=p[i-1]+1;
        else p[i]=p[i-1];
        cout<<i<<" "<<p[i]<<"  "<<(i-4)/2<<" "<<(int)sqrt(i*1.0)<<" "<<endl;
    }
    return 0;
}