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題目大意：給兩個(gè)數(shù)a，b，求滿足c*d==a且c>=b且d>=b的c,d二元組對(duì)數(shù)，(c,d)和(d,c)屬于同一種情況 題目分析：根據(jù)唯一分解定理先將a唯一分解，則a的所有正約數(shù)的個(gè)數(shù)為ans?= (1 + a1) * (1 + a2) *...(1 + an) 因?yàn)轭}目說(shuō)了不會(huì)存在c==d的情況，因此ans要除2，去掉重復(fù)情況。[care] 然后枚舉小于b的a的約數(shù)，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(shù)基本原理:任何一個(gè)大于1的自然數(shù),都可以唯一分解成有限個(gè)質(zhì)數(shù)的乘積 N=p1^a1*p2^a2.....pn^an,這里p1<p2<...<pn均為質(zhì)數(shù)，其諸指數(shù)是正整數(shù)。 定理應(yīng)用：（1）一個(gè)大于1的正整數(shù)N，如果它的標(biāo)準(zhǔn)分解式為：N=p1^a1*p2^a2.....pn^an， 那么它的正因數(shù)個(gè)數(shù)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?yàn)槿魏我粋€(gè)整數(shù)都可以由一個(gè)素?cái)?shù)經(jīng)過(guò)乘法運(yùn)算得到，故可通過(guò)素?cái)?shù)打表的方式求得某數(shù)的唯一分解（得到冪次最大）*/if(!book[i]){dp[k++]=i;/*記錄素?cái)?shù)*/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(dǎo)致錯(cuò)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(duì)m的值進(jìn)行操作，得到x=a1^b1*a2^b2.....an^bn*/a++;}ans=ans*(a+1);/**a的所有正約數(shù)的個(gè)數(shù)為ans = (1 + a1) * (1 + a2) *...(1 + an)*/}if(m==1)break;}if(m>1)/**for循環(huán)，先判斷，導(dǎo)致停止，故對(duì)最后出現(xiàn)的一個(gè)素?cái)?shù)的情況判斷*2*/ans*=2;ans/=2;///因?yàn)轭}目說(shuō)了不會(huì)存在c==d的情況，因此ans要除2，去掉重復(fù)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(shù)對(duì)個(gè)數(shù)時(shí)，出現(xiàn)了1*m的情況（當(dāng)所有素?cái)?shù)都為1時(shí)，該情況出現(xiàn)），此時(shí)減去*/if(mm%i==0)/**care m的值在上面操作中發(fā)生改變，故需要開(kāi)變量存儲(chǔ)m值*/b++;ans-=b;/**題目要求，不出現(xiàn)小于n的因數(shù)對(duì)，找出減去即可*/}printf("Case %d: %lld\n",tt++,ans);}return 0; }

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