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題目大意：給兩個數a，b，求滿足c*d==a且c>=b且d>=b的c,d二元組對數，(c,d)和(d,c)屬于同一種情況 題目分析：根據唯一分解定理先將a唯一分解，則a的所有正約數的個數為ans?= (1 + a1) * (1 + a2) *...(1 + an) 因為題目說了不會存在c==d的情況，因此ans要除2，去掉重復情況。[care] 然后枚舉小于b的a的約數，拿ans減掉就可以了

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

算數基本原理:任何一個大于1的自然數,都可以唯一分解成有限個質數的乘積 N=p1^a1*p2^a2.....pn^an,這里p1<p2<...<pn均為質數，其諸指數是正整數。 定理應用：（1）一個大于1的正整數N，如果它的標準分解式為：N=p1^a1*p2^a2.....pn^an， 那么它的正因數個數為f(n)=(1+a1)(1+a2).....(1+an)。

AC代碼分步詳解

#include<iostream> #include<string.h> #include<math.h> typedef long long ll; #include<stdio.h> using namespace std; #define M 1000010 int dp[M]; int book[M]; int t,k; ll m,n; void dfs() {k=0;memset(dp,0,sizeof(dp));memset(book,0,sizeof(book));for(int i=2; i<M; i++)/**因為任何一個整數都可以由一個素數經過乘法運算得到，故可通過素數打表的方式求得某數的唯一分解（得到冪次最大）*/if(!book[i]){dp[k++]=i;/*記錄素數*/for(int j=2*i; j<M; j+=i)book[j]=1;} } int main() {cin>>t;int tt=1;dfs();/*最好放在外面*/while(t--){cin>>m>>n;ll ans=1;/**定義為 long long 型，否則導致錯誤答案*/ll mm=m;if(n>sqrt(m))ans=0;else{for(int i=0; i<k&&2*dp[i]<m ; i++) /**care:remember停止條件為i<k&&2*dp[i]<=m，缺一不可*/{if(m%dp[i]==0){int a=0;while(m%dp[i]==0){m/=dp[i];/**唯一分解定理：直接對m的值進行操作，得到x=a1^b1*a2^b2.....an^bn*/a++;}ans=ans*(a+1);/**a的所有正約數的個數為ans = (1 + a1) * (1 + a2) *...(1 + an)*/}if(m==1)break;}if(m>1)/**for循環，先判斷，導致停止，故對最后出現的一個素數的情況判斷*2*/ans*=2;ans/=2;///因為題目說了不會存在c==d的情況，因此ans要除2，去掉重復情況。int b=0;for(ll i=1; i<n; i++)/**i=1,我的理解是前面由唯一分解定理求因數對個數時，出現了1*m的情況（當所有素數都為1時，該情況出現），此時減去*/if(mm%i==0)/**care m的值在上面操作中發生改變，故需要開變量存儲m值*/b++;ans-=b;/**題目要求，不出現小于n的因數對，找出減去即可*/}printf("Case %d: %lld\n",tt++,ans);}return 0; }

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