LightOJ 1336(Sigma Function)

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Sigma Function

Sigma function is an interesting function in Number Theory. It is denoted by the Greek letter Sigma (σ). This function actually denotes the sum of all divisors of a number. For example σ(24) = 1+2+3+4+6+8+12+24=60. Sigma of small numbers is easy to find but for large numbers it is very difficult to find in a straight forward way. But mathematicians have discovered a formula to find sigma. If the prime power decomposition of an integer is

 

Then we can write,

 

For some n the value of σ(n) is odd and for others it is even. Given a value n, you will have to find how many integers from 1 to n have even value of σ.


Input

Input starts with an integer T (≤ 100), denoting the number of test cases.

Each case starts with a line containing an integer n (1 ≤ n ≤ 1012).

Output

For each case, print the case number and the result.

Sample Input

4

3

10

100

1000

Sample Output

Case 1: 1

Case 2: 5

Case 3: 83

Case 4: 947

 

题意:函数σ(n)等于n的所有因数之和,问满足σ(k)为偶数的k(1<=k<=n)有多少个。

分析:题目给出求σ(n)的公式σ(n)=((p1^(e1+1)-1)/(p2-1))*((p2^(e2+1)-1)/(p2-1))*......*((pk^(ek+1)-1)/(pk-1)),

又根据等比数列公式(p^(e+1)-1)/(p-1)=1+p^1+p^2+......+p^e,

我们可以先求σ(k)为奇数的个数ans再用n减去ans便是σ(k)为偶数的个数:

σ(n)为奇数时,(p^(e+1)-1)/(p-1)=1+p^1+p^2+......+p^e 必为奇数,

(1) 当p为偶数的时候,因为既是偶数又是素数的只有2,故p=2,

此时(p^(e+1)-1)/(p-1)=2^(e+1)-1必为奇数;

(2)当p为奇数时,p^k必为奇数,要使(p^(e+1)-1)/(p-1)=1+p^1+p^2+......+p^e为奇数,

e+1必为奇数,则e必为偶数;

因此,只有p1,p2,......pk满足(1)或(2)的时候σ(k)为奇数。

由算术基本定理知,n=p1^e1 * p2^e2 * ...... * pk^ek

1.当n满足条件(2)时,因为ei为偶数,所以n必为完全平方数,而且可以证明完全平方数n的σ(n)必为奇数;

2.当n满足条件(1)时,n=2^t * p1^e1 * p2^e2 * ...... * pk^ek ,若t为偶数则它n就是完全平方数,即上面的1,

若t为奇数则n=2*X (X为完全平方数)。

因此,只要排除掉n以内的所有完全平方数X和2*X,就可以得到σ(k)为偶数的个数。

 

技术分享图片
#include<cstdio>
#include<cmath>
int main()
{
    int T,cas=0;
    long long N,ans;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%lld",&N);
        ans=(long long)sqrt(N)+(long long)sqrt(N/2.0);
        printf("Case %d: %lld\n",++cas,N-ans);
    }
    return 0;
}
View Code

 

 

 

 

 

 

 


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