Light OJ Aladdin and the Flying Carpet(约数个数)

Input

Input starts with an integer T (≤ 4000), denoting the number of test cases.

Each case starts with a line containing two integers: ab(1 ≤ b ≤ a ≤ 10¹²) wherea 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

大致题意：有T（<=4000）组数据，求面积是a且边长均不小于b的矩形个数

思路:直接枚举b至sqrt（a）的约数分数会超时，然而先求出总的约数个数/2（成对出现)再减去小于b的所有约数即可（说明b是总体偏小的）

#include<cstdio>#include<iostream>#include<cstring>#include<algorithm>using namespace std;typedef long long ll;const ll MAXN = 1e6+1000;bool vis[MAXN];ll prime[MAXN];int main(){    int cnt=0;    for(int i=2;i*i<MAXN;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])  prime[cnt++]=i;    int T;    scanf("%d",&T);    for(int cas=1;cas<=T;cas++)    {        int ans=0;        ll s,lim;        cin>>s>>lim;        ll tmp=s;        if(s>lim*lim)        {            ans=1;            for(int i=0;prime[i]*prime[i]<=s;i++)            {                if(s%prime[i]) continue;                int tmp=0;                while(s%prime[i]==0)                {                    tmp++;                    s/=prime[i];                }                ans*=(tmp+1);            }            if(s>1) ans*=2;            ans>>=1;            for(int i=1;i<lim;i++) if(tmp%i==0) ans--;        }        cout<<"Case "<<cas<<": "<<ans<<endl;    }    return 0;}