Aladdin and the Flying Carpet

Aladdin and the Flying Carpet

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Time Limit: 3 second(s)Memory Limit: 32 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 strange magicalflying carpet at the entrance. There were some strange creatures guarding theentrance of the cave. Aladdin could run, but he knew that there was a highchance of getting caught. So, he decided to use the magical flying carpet. Thecarpet was rectangular shaped, but not square shaped. Aladdin took the carpetand with the help of it he passed the entrance.

Now you are given the area of the carpet and the length ofthe minimum possible side of the carpet, your task is to find how many types ofcarpets are possible. For example, the area of the carpet 12, and the minimumpossible side of the carpet is 2, then there can be two types of carpets andtheir 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:ab (1 ≤ b ≤ a ≤ 10¹²) where adenotes the area of the carpet andb denotes the minimum possible sideof the carpet.

Output

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

输入：

2

10 2

12 2

输出：

Case 1: 1

Case 2: 2

题意：

给定T组数据，每组数据有两个数 面积s 和 矩形的宽 a，让你求的是在面积s一定的情况下，假设长和宽分别为d和 e,最小的边长 >= a的有几种方式可以组成矩形（不是正方形）

解析一下样例：

面积为12 ，最小的边长为2：

首先将12进行素因子分解12 = 2^2*3，所以我们能够得到 

12 = 1 * 12（不符合条件 最小的边长<2）

12 = 2 * 6（符合）

12 = 3 * 4（符合）

所以有 2 种方式，输出 2

解题思路：

每次做题的时候先要考虑一下能不能暴力（暴力简单），这个题如果我们要暴力的话肯定会超时，所以就不要暴力了

通过上述的样例分析，我们也知道了我们首先要做的就是将 面积 s 进行素因子分解，这就用到了唯一分解定理，

s = p1^e1 * p2^e2 *……* pk^ek，我们要得到的是因子的个数，这里所说的因子个数默认为正的，得到因子个数的方法是 num = (e1+1) * (e2+1) * ... *(ek+1)，然后又因为没有正方形，而且我们要得到的是有多少对，所以将
num除以2，就得到了可以组成矩形面积为 s 的矩形个数，然后我们只需要在 [1,a)的区间内（注意区间开闭）将 s 的因子减掉就行了（num--），这样就可以了。

科普一下唯一分定理：

定义：任何一个大于1的自然数N,如果N不为质数，那么N可以唯一分解成有限个质数的乘积N=这里P_1<P_2<P_3......<P_n均为质数，其中指数a_i是正整数。这样的分解称为N的标准分解式

应用：

1、当一个数太大long long无法表示,而且需要和其他数约分时可以使用唯一分解定理uva10375

2、最小公倍数uva10791

3、小于n且与n互素的整数个数：欧拉函数

#include <iostream>#include <cmath>#include <cstring>#include <algorithm>#include <cstdio>using namespace std;typedef long long LL;const int MAXN = 1e6+5;bool prime[MAXN];LL p[MAXN],k;void isprime()//筛法素数{    memset(prime, false, sizeof(prime));    prime[1] = true;    k = 0;    for(LL i=2; i<MAXN; i++)    {        if(!prime[i])        {            p[k++] = i;            for(LL j=i*i; j<MAXN; j+=i)                prime[j] = true;        }    }}LL Solve(LL m)//找这个数的因子的个数{    LL ret = 1;    for(LL i=0; p[i]*p[i]<=m&&i<k; i++)    {        LL cnt = 0;        if(m%p[i] == 0)        {            while(m%p[i] == 0)            {                cnt++;                m /= p[i];            }            ret *= (cnt+1);        }    }    if(m > 1)        ret *= 2;    return ret;}int main(){    isprime();    int T;    cin>>T;    for(int cas=1; cas<=T; cas++)    {        LL s,a;        cin>>s>>a;        if(a*a >= s)            printf("Case %d: 0\n",cas);        else        {            LL ret = Solve(s);            ret /= 2;//因为求得是成对的因子，，所以总的因子个数/2            for(LL i=1; i<a; i++)//因为要求两个因子都大于等于a。。所以把s在1~a范围内的因子除去就行                if(s % i == 0)                    ret--;            printf("Case %d: %lld\n",cas,ret);        }    }    return 0;}