LightOJ1336 Sigma Function (number of divisors and odd numbers) __ Math-theory/game

F (n) is the and of all the divisors of N, give you a number of n, let you ask from 1 to n in the number of F (n) are even numbers how many

Analysis:

The factor of number x and f (x) = (1+P1+P1^2+P1^3+...+P1^A1) * (1+P2+P2^2+...+P2^A2) *...* (1+pn+pn^2+...+pn^an);

Because even or even, odd-numbered or odd-numbered, odd-numbered even is even, all must have an odd number in each bracket, and then minus the divisors and odd numbers is the answer.

1. When x is 2, 2 of the corresponding brackets and must be an odd number, because even plus 1 must be odd.

2. Except for 2, all primes are odd, and to make x the corresponding number of other elements of the parentheses and the odd, you must ensure that X has even several of the factor, that AI must be even.

3. The number that satisfies the above two conditions is a square number, that is, the divisor and the odd number x, it must be a square number, and of course the number x multiplied by 2 is also satisfying the 2*x and odd numbers.

So just subtracting the n minus sqrt (n) and sqrt (N/2) is the answer. You can also find the number of squares within n, and the number of 2* squared number not more than n, and the answer is after the n minus.

The first method:

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <cmath>
using namespace std;
typedef long long LL;
int main ()
{
    int T;
    scanf ("%d", &t);
    for (int kase = 1; kase <= T; kase++)
    {
        ll n, sum;
        scanf ("%lld", &n);
        sum = n;
        Sum-= (int) sqrt (n);
        Sum-= (int) sqrt (N/2);
        printf ("Case%d:%lld\n", Kase, sum);
    }
    return 0;
}

The second method:

#include <stdio.h>
int main ()
{
    int T, Kase = 0;
    Long long N;
    scanf ("%d", &t);
    while (t--)
    {
        scanf ("%lld", &n);
        Long ans = 0;
        for (Long i = 1; i*i <= N; i++)
        {
            ans++;
            if (2*i*i<=n)
                ans++;
        }
        printf ("Case%d:%lld\n", ++kase, N-ans);
    }
    return 0;
}