C - Aladdin and the Flying Carpet

最新推荐文章于 2021-06-04 20:40:21 发布

转载最新推荐文章于 2021-06-04 20:40:21 发布 · 78 阅读

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原文链接：http://www.cnblogs.com/joeylee97/p/6826183.html

本文讲述了Aladdin使用神奇飞毯通过魔法洞穴入口的故事，并以此为背景，介绍了一个数学问题：如何根据飞毯的面积和最小边长，计算可能的飞毯类型数量。文章提供了一段C++代码，利用唯一分解定理来解决这一问题，通过寻找所有质因数并排除小于等于特定值的情况，从而得出答案。

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 bdenotes 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

唯一分解定理

#include<iostream>
#include<cstdio>
#include<cstring>
#include<sstream>
#include<algorithm>
#include<queue>
#include<vector>
#include<cmath>
#include<map>
#include<stack>
#include<set>
#include<fstream>
#include<memory>
#include<list>
#include<string>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
#define MAXN  1000010
#define LLL 1000000000
#define INF 1000000009
/*
数学唯一分解定理
求出所有a的素因子 然后减去小于等于x的
*/
vector<int> prime, v;
bool notprime[MAXN];
void init()
{
    memset(notprime, false, sizeof(notprime));
    notprime[0] = notprime[1] = true;
    for (int i = 2; i < MAXN; i++)
    {
        if (!notprime[i])
        {
            prime.push_back(i);
            if (i > MAXN / i) continue;
            for (int j = i*i; j < MAXN; j += i)
            {
                notprime[j] = true;
            }
        }
    }
}
LL count(LL x)
{
    LL ans = 1;
    for (int i = 0; i < prime.size()&&prime[i]<=sqrt(x*1.0); i++)
    {
        LL tmp = 0;
        while (x%prime[i] == 0)
        {
            tmp++;
            x /= prime[i];
        }
        ans *= (tmp + 1);
    }
    if (x > 1)
        ans *= 2;
     return ans;
}
int main()
{
    init();
    int T;
    LL n, x;
    scanf("%d", &T);
    for(int cas=1;cas<=T;cas++)
    {
        scanf("%lld%lld", &n, &x);
        if (x*x >= n)
        {
            printf("Case %d: %d\n", cas, 0);
            continue;
        }
        LL ans = count(n)/2;
        LL cnt = 0;
        for (LL i = 1; i<x; i++)
        {
            if (n%i == 0)
                cnt++;
        }
        printf("Case %d: %lld\n",cas, ans - cnt);
    }
    return 0;
}

转载于:https://www.cnblogs.com/joeylee97/p/6826183.html