LightOJ 1336 Sigma Function 数论 有意思

题目连接 1336 - Sigma Function ---------------------

 

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Time Limit: 2 second(s)

Memory Limit: 32 MB

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

\[n={ { p }_{ 1 } }^{ { e }_{ 1 } }\cdot { { p }_{ 2 } }^{ { e }_{ 2 } }\cdot { { p }_{ 3 } }^{ { e }_{ 3 } }...{ { p }_{ k } }^{ { e }_{ k } }\]

Then we can write,

\[\sigma (n)=\frac { { { p }_{ 1 } }^{ { e }_{ 1 }+1 }-1 }{ { p }_{ 1 }-1 } \cdot \frac { { { p }_{ 2 } }^{ { e }_{ 2 }+1 }-1 }{ { p }_{ 2 }-1 } ...\frac { { { p }_{ k } }^{ { e }_{ k }+1 }-1 }{ { p }_{ k }-1 } \]

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

Output for Sample Input

4

3

10

100

1000

Case 1: 1

Case 2: 5

Case 3: 83

Case 4: 947

题意：

定义\(\sigma (n)\)为\(n\)的因子和，求1～\(n\)范围内\(\sigma\)值为偶数的有多少个。

思路：

首先分解\(n\), $n={ i=1 }^{ r }{ { { p }{ i } }^{ { a }{ i } } } \(,则\)(n)={ i=1 }^{ r }{ ({ j=0 }^{ { a }{ i } }{ { { p }{ i } }^{ j } } ) } \(，我们考虑什么情况下\)(n)$为奇数，有 * \({ p }_{ i }=2\)时，${ j=0 }^{ { a }{ i } }{ { { p }{ i } }^{ j } } $为奇数 * \({ p }_{ i }\neq 2\)时，${ a }{ i } $ 均为偶数时 ${ j=0 }^{ { a }{ i } }{ { { p }{ i } }^{ j } } \(为奇数 因此，我们发现\){ a }{ i } \(均为偶数时，有\) ={ i=1 }^{ r }{ { { p }_{ i } }^{  } } \(为因子和为奇数的个数，但是缺少了2的幂为奇数的情况，共有\) \(种情况，因此最后的答案为\)n- - $

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#define LL long long
using namespace std;
int main()
{
    int T;
    scanf("%d",&T);
    for(int cas=1;cas<=T;cas++)
    {
        LL n;
        scanf("%lld",&n);
        LL ans =((LL)sqrt(n)+(LL)sqrt(n/2.));
        printf("Case %d: %lld\n",cas,n-ans);
    }
    return 0;
}