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
题目链接：http://lightoj.com/login_main.php?url=volume_showproblem.php?problem=1336
题意：f(n)为n所有约数的和，给你一个数n,让你求从1到 n 中 由多少个f(x)的值为偶数。
将公式化简成 f(n) = （1+p1+p1^2+p1^3+...+p1^a1）*(1+p2+p2^2+...+p2^a2)*...*(1+pn+pn^2+...+pn^an)。
偶数*偶数==偶数，奇数*奇数 == 奇数，奇数*偶数 == 偶数 奇数or偶数＋1 = 偶数or奇数
由此可以看出奇数项会比较少，所以说答案 = n - 奇数项即可。
因为素数里面除了2就都是奇数，因此上式中第一项一定是奇数，要使其他的项是奇数（当且仅当ai = 偶数）。因此满足奇数项的就是 sqrt（n）+sqrt(n/2)
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<cmath>
using namespace std;
typedef long long ll;
int main()
{
int t;
cin >> t;
int cas = 1;
while(t --)
{
ll n, sum;
cin >> n;
sum = n;
sum -= (int)sqrt(n);
sum -= (int)sqrt(n/2);
printf("Case %d: %lld\n", cas++, sum);
}
return 0;
}