Aladdin and the Flying Carpet 唯一分解定理

Description

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

2
10 2
12 2

Sample Output

Case 1: 1
Case 2: 2

Hint

题意

给一个矩形的面积 求长和宽 不能为正方形 给定限制b要求最小边大于等于b 求满足长和宽个数

题解：

使用唯一分解定理求解

AC代码

#include <cstdio>#include <cstring>typedef long long ll;#define N 1000005int  prime[N];bool vis[N];ll cnt ;void set_prime(){    cnt = 0;    memset(vis,false,sizeof(vis));    for (int i = 2; i < N; ++i){        if (!vis[i]) {            prime[cnt++] =  i;            for (int j = i; j < N; j+=i){            vis[j] = true;            }        }    }}ll set_solve(ll n){    ll ck;    ll ans = 1;    for (int i = 0; i < cnt&&prime[i]<=n; ++i){        if (n%prime[i]==0){            ck = 0;            while(n%prime[i]==0){                n/=prime[i];ck++;            }            ans *= (ck+1);        }    }    if  (n>1) ans*=2;  //n为质数 情况只有2种（含重复）    return ans;}int main(){    int  t,kase = 1;ll a,b;    set_prime();    scanf("%d",&t);    while (t--){        scanf("%lld%lld",&a,&b);        if(a/b < b)        {           printf("Case %d: %lld\n",kase++,0);            continue;        }        ll as = set_solve(a);        as/=2; //除去重复的        for (int i = 1; i < b; ++i){            if (a%i==0) as--;        }//再减去不满足的        printf("Case %d: %lld\n",kase++,as);    }    return 0;}