1341 - Aladdin and the Flying Carpet

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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.

题意：

给一对数字 a，b  ，a是一个长方形的面积，问有多少种整数的边的组合可以组成面积为a的长方形，要求最短的边不得小于b。

说白了，就是求区间[b, a] 内的 a 的约数对的个数。约数对就是【4,5】【5,4】这就是一对。救赎不考虑顺序。

解题思路：

主要利用算术基本定理，先介绍下算数基本定理几个性质：

（1）一个大于1的正整数N，如果它的标准分解式为：

，那么它的正因数个数为

。

（2） 它的全体正因数之和为

。

当

时就称N为完全数。 是否存在奇完全数，是一个至今未解决之猜想。

（3） 利用算术基本定理可以重新定义整数a和b的最大公因子

和最小公倍数

， 并证明

。

（4）此外还可证明根号2是无理数等等。

（5）证明素数个数无限。

这个题主要就是求正因子个数num，求出来之后num/2，然后在枚举1-b 中属于a正因子的数cnt， 用num-cnt就是我们最后要的结果。

在处理正因子的时候要注意。并且，在num除2的时候，我们会想到num为奇数的时候直接除2会不会出错？其实刚好，只有当有一个约数所形成的边是正方形的时候，num那一定是奇数，直接/2刚好减去了正方形的情况。

AC代码：

#include <stdio.h>
#include <math.h>
#include <vector>
#include <queue>
#include <string>
#include <string.h>
#include <stdlib.h>
#include <iostream>
#include <algorithm>

using namespace std;

const int MAXN = 1000000;
int prime[MAXN];
int is_prime[MAXN];
int tot;

typedef long long LL;

void find_prime()
{
    tot=0;
    is_prime[1]=1;
    for(int i=2;i<MAXN;i++){
        if(!is_prime[i]){
            prime[tot++]=i;
            for(int j=i*2;j<MAXN;j+=i) is_prime[j]=1;
        }
    }
}

LL solve(LL n)
{
    LL ans=1;
    for(int i=0;i<tot&&n;i++){
        LL num=0;
        if(prime[i]>n) break; //不加这个超时啊。。。
        while(n%prime[i]==0){
            n/=prime[i];
            num++;
        }
        ans*=(1+num);
    }
    if(n>1) ans*=(1+1);
    return ans;
}

int main()
{
    find_prime();
    LL s,b;
    int t;
    scanf("%d",&t);
    int xp=1;
    while(t--){
        scanf("%lld%lld",&s,&b);
        if(b*b>=s){
        	printf("Case %d: 0\n",xp++);
        	continue;
		}
		LL num=solve(s);
        num/=2;
        for(LL i=1;i<b;i++)
            if(s%i==0) num--;
        printf("Case %d: %lld\n",xp++,num);
    }
    return 0;
}

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