数论 C - Aladdin and the Flying Carpet

最新推荐文章于 2020-03-05 17:41:59 发布

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最新推荐文章于 2020-03-05 17:41:59 发布

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原文链接：http://www.cnblogs.com/EchoZQN/p/10753001.html

本文探讨了一个基于唯一分解定理的数学问题，通过一个类似于《阿拉丁》故事的情境，解释了如何计算一个矩形飞毯可能的尺寸组合，前提是已知飞毯的面积和最小边长。文章详细介绍了使用C++实现的算法，包括初始化素数表、计算正因数数量以及最终确定符合条件的飞毯类型。

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It's said that Aladdin had to solve seven mysteries before getting the Magical Lamp which summons a powerful Genie. Here we are concerned about the first mystery.

Aladdin was about to enter to a magical cave, led by the evil sorcerer who disguised himself as Aladdin's uncle, found a strange magical flying carpet at the entrance. There were some strange creatures guarding the entrance of the cave. Aladdin could run, but he knew that there was a high chance of getting caught. So, he decided to use the magical flying carpet. The carpet was rectangular shaped, but not square shaped. Aladdin took the carpet and with the help of it he passed the entrance.

Now you are given the area of the carpet and the length of the minimum possible side of the carpet, your task is to find how many types of carpets are possible. For example, the area of the carpet 12, and the minimum possible side of the carpet is 2, then there can be two types of carpets and their sides are: {2, 6} and {3, 4}.

Input

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

Each case starts with a line containing two integers: a b (1 ≤ b ≤ a ≤ 10¹²) where a denotes the area of the carpet and b denotes the minimum possible side of the carpet.

Output

For each case, print the case number and the number of possible carpets.

Sample Input

10 2

12 2

Sample Output

Case 1: 1

Case 2: 2

这个题目很明显是唯一分解定理，但是如果你不知道唯一分解定理，那这个其实就有点难。

如果明白这个这个定理，那么这个题目就变得很容易了，这个题目就是运用了这个定理。

题目让你求一个数的分解形式有多少种，且分解成的最小的数要比给定数字大，

那不就是你求出有多少个正因数，然后除以2，这个求的就是对数。

然后减去不满足条件的。

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <queue>
#include <vector>
#include <cmath>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long ll;
const int maxn = 1e6;
int v[maxn],isp[maxn], m;

void init1()
{
    m = 0;
    memset(v, 0, sizeof(v));
    for (int i = 2; i < maxn; i++)
    {
        if (v[i] == 0)
        {
            isp[m++] = i;
            v[i] = i;
        }
        for (int j = 0; j < m; j++)
        {
            if (v[i]<isp[j] || i * isp[j]>maxn) break;
            v[i*isp[j]] = isp[j];
        }
    }
}


ll cont(ll x)
{
    ll sum = 1;
    if (x == 0) return 0;
    for(ll i=0;i<m;i++)
    {
        ll num = 0;
        while(x%isp[i]==0)
        {
            x /= isp[i];
            num++;
        }
        sum *= num + 1;
        if (x == 1) break;
    }
    if (x > 1) sum *= 2;
    return sum;
}

int main()
{
    int t, cas = 0;
    init1();
    cin >> t;
    while(t--)
    {
        ll a, b;
        cin >> a >> b;
        if(b>=sqrt(a))
        {
            printf("Case %d: %d\n", ++cas, 0);
        }
        else
        {
            ll cnt = 0;
            for(ll i=1;i<b;i++)
            {
                if (a%i == 0) cnt++;
            }
            ll sum = cont(a) / 2;
            ll ans = sum - cnt;
            printf("Case %d: %lld\n", ++cas, ans);
        }
    }
    return 0;
}