Lightoj 1341 Aladdin and the Flying Carpet (unique decomposition theorem)

Aladdin and the Flying Carpet

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Time Limit: 3 second (s)

Memory Limit: MB

It's said that Aladdin had to solve seven mysteries beforegetting the magical Lamp which summons a powerful Genie. Here we are concernedabout the first mystery.

Aladdin was-about-to-enter to a magical-cave, led by Theevil sorcerer who disguised himself as Aladdin ' s uncle, found a St Range Magicalflying carpet at the entrance. There were some strange creatures guarding theentrance of the cave. Aladdin could run, but he knew the there was a highchance of getting caught. So, he decided to use the magical flying carpet. Thecarpet was rectangular shaped and not square shaped. Aladdin took the carpetand with the help of it he passed the entrance.

Now all given the area of the carpet and the length ofthe minimum possible side of the carpet, your task was to find Ho W Many types ofcarpets is possible. For example, the area of the carpet, and the minimumpossible side of the carpet are 2, then there can be both types of CA Rpets Andtheir sides is: {2, 6} and {3, 4}.

Input

Input starts with an integer T (≤4000), denoting the number of test cases.

Starts with a line containing-integers:ab (1≤b≤a≤1012) where adenotes the area of th E Carpet andb denotes the minimum possible sideof the carpet.

Output

For each case, print the case number and the number ofpossible carpets.

Sample Input	Output for Sample Input
2 10 2 12 2	Case 1:1 Case 2:2

Title Link: http://lightoj.com/volume_showproblem.php?problem=1341

Title: To two number A, a, to satisfy C*d==a and c>=b and d>=b c,d two-tuple logarithm, (c,d) and (D,c) belong to the same situation

Title Analysis: According to the unique decomposition theorem, a unique decomposition, then the number of all positive numbers of a is num = (1 + A1) * (1 + a2) * ... (1 + AI), where the AI is the exponent of the element factor, see the unique decomposition theorem, because the topic says there will be no c==d, so num to remove 2, eliminate the repetition, and then enumerate the approximate number of a less than B, take Num to lose it.

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#define ll long long
using namespace std;
int const MAX = 1e6 + 10;
int p[MAX];
bool u[MAX];
int num, cnt;
ll a, b, tmp;

void get_prime()  
{  
    memset(u, false, sizeof(u));
    for(int i = 2; i <= sqrt(MAX); i++)
        if(!u[i])
            for(int j = i * i; j <= MAX; j += i)
                u[j] = true;
    for(int i = 2; i <= MAX; i++)
        if(!u[i])
            p[cnt ++] = i;
}

void cal()
{
    for(int i = 0; i < cnt && p[i] <= sqrt(tmp); i++)
    {
        int cc = 0;
        while(tmp % p[i] == 0)
        {
            cc ++;
            tmp /= p[i];
        }
        num *= (cc + 1);

    }
    if(tmp > 1) //如果tmp不能被整分，说明还有一个素数是它的约数，此时cc=1
        num *= 2;
}

int main()
{
    int T;
    scanf("%d", &T);
    cnt = 0;
    get_prime();
    for(int ca = 1; ca <= T; ca++)
    {
        scanf("%lld %lld", &a, &b);
        if(a < b * b)
            printf("Case %d: 0\n", ca);
        else
        {
            num = 1;
            tmp = a;
            cal();
            num /= 2;
            for(int i = 1; i < b; i++)
                if(a % i == 0)
                    num --;
            printf("Case %d: %d\n", ca, num);
        }
    }
}

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Lightoj 1341 Aladdin and the Flying Carpet (unique decomposition theorem)