LightOj 1341 Aladdin and the Flying Carpet (素数筛+求一个数的约数个数)

传送门：http://lightoj.com/volume_showproblem.php?problem=1341

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 adenotes 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

2

10 2

12 2

Sample Output

Case 1: 1

Case 2: 2

题目大意：

给出两个个数a和b，求有多少对{（h,w）| h>=b&&w>=b }使得h*w=a。这个题如果不失一般性的话暴力和其他的办法所消耗的最大时间是一样的，都是O(T*sqrt(a))，但是就是卡数据。

题解：

先预处理求1e6以内的所有素数，然后求a的约数个数Num=（a=p1^a1 * p2^a2 * p3^a3 * ......）=（a1+1）*(a2+1)*(a3+1)*..........，因为(w,h)和（h,w）算一对，所以num/2;然后在枚举小于b的减去（其实如果正常的话在枚举<b的时候的复杂度也是1e6）。

///http://lightoj.com/volume_showproblem.php?problem=1341#include <iostream>#include <cstring>#include <cstdio>#include <cstdlib>#include <algorithm>#include <cmath>using namespace std;typedef long long LL;const int MAXN=1e6+7;int top,vis[MAXN+100],pri[MAXN+100];void getprime(){    top=0;    memset(vis,0,sizeof(vis));    for(int i=2; i<=sqrt(MAXN+0.5); i++)    {        if(!vis[i])        {            for(int j=i*i; j<=MAXN; j+=i)                vis[j]=1;        }    }    for(int i=2; i<=MAXN; i++)    {        if(!vis[i])pri[++top]=i;    }}int main(){    getprime();    int T,w=0;    LL a,b;    scanf("%d",&T);    while(T--)    {        scanf("%lld%lld",&a,&b);        if(a<b*b)        {            printf("Case %d: 0\n",++w);            continue;        }        LL num=1,temp=a;        for(int i=1; i<=top&&pri[i]<=sqrt(temp); i++)        {            if(temp%pri[i]==0)            {                int k=0;                while(temp%pri[i]==0)temp/=pri[i],k++;                num=num*(k+1);            }        }        if(temp>1)num=num*2;        num=num/2;        for(int i=1; i<b; i++)if(a%i==0)num--;        printf("Case %d: %lld\n",++w,num);    }    return 0;}