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

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

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=\sum _{ i=1 }^{ r }{ { { p }_{ i } }^{ { a }_{ i } } } $,则$\sigma (n)=\prod _{ i=1 }^{ r }{ (\sum _{ j=0 }^{ { a }_{ i } }{ { { p }_{ i } }^{ j } } ) } $,我们考虑什么情况下$\sigma (n)$为奇数,有

  • ${ p }_{ i }=2$时,$\sum _{ j=0 }^{ { a }_{ i } }{ { { p }_{ i } }^{ j } } $为奇数
  • ${ p }_{ i }\neq 2$时,${ a }_{ i } $ 均为偶数时 $\sum _{ j=0 }^{ { a }_{ i } }{ { { p }_{ i } }^{ j } } $为奇数
    因此,我们发现${ a }_{ i } $均为偶数时,有$\sqrt { n } =\sum _{ i=1 }^{ r }{ { { p }_{ i } }^{ \frac { { a }_{ i } }{ 2 } } } $为因子和为奇数的个数,但是缺少了2的幂为奇数的情况,共有$\sqrt { \frac { n }{ 2 } } $种情况,因此最后的答案为$n-\sqrt { n } -\sqrt { \frac { n }{ 2 } } $

代码:

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#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;
}