Aladdin and the Flying Carpet（算术基本定理）
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: ab(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
题意：给个矩形的面积a,和矩形的最小边长b，问有多少种矩形的方案(不能是正方形)
分析：a可以写成x,y，因为不能是正方形，所以设x<y，那么x<sqrt(a),y>sqrt(a)
所以找到所有小于sqrt(a)的因子，看有几个大于等于b的就是方案数
因子可以由数的唯一分解定理，求得
具体 ： 先筛一遍1e6以内的素数，有线性筛，然后分解a,然后dfs找所有的小于sqrt(a)的因子，
由于前12个素数的乘积大于1e12了，所以这部分复杂度，大概是O（2^12）（一般还要略大，不过大不了多少，数组要开大）左右
可以用这个估计(因为是求小于sqrt(a)的，可以除以2，当然这是空间常数）
所以这部分复杂度是O（T*2^12)满的话（4000*4000）大概也就是几百万，这部分可以忽略不计
主要的复杂度在分解素数里，因为1e6里面大概有7w多素数，这部分复杂度（最坏的情况a是大素数），大概是4000*70000，可以卡过，由于不可能都是这种数据
所以还是可以过的
吐槽：然后我看了看网上的代码，都是先求出总的，然后暴力扫b减，结果居然过了，b是sqrt(a)的级别，是百万，4000*1e6,是4e9，TLE
出题人太良心，没有卡这种的QAQ，感觉略坑啊
#include <cstdio>
#include <iostream>
#include <ctime>
#include <vector>
#include <cmath>
#include <map>
#include <queue>
#include <algorithm>
#include <cstring>
using namespace std;
typedef long long LL;
const int N=1e6+5;
const int INF=0x3f3f3f3f;
int cnt;
bool v[N];
LL prime[80000];
void getprime()
{
for(int i=2; i*i<=N-5; ++i)
if(!v[i])
for(int j=i*i; j<=N-5; j+=i)
v[j]=1;
for(int i=2; i<=N-5; ++i)
if(!v[i])
prime[++cnt]=i;
}
vector<LL>fac[2];
int divisors[5000],tot;
LL k;
void dfs(int pos,LL res)
{
if(pos==fac[0].size())
{
divisors[++tot]=res;
return;
}
for(LL i=0,now=1; i<=fac[1][pos]; now*=fac[0][pos],++i)
{
if(now*res>=k)
break;
dfs(pos+1,res*now);
}
}
int main()
{
getprime();
int cas=0,T;
scanf("%d",&T);
while(T--)
{
printf("Case %d: ",++cas);
LL a,b;
scanf("%lld%lld",&a,&b);
k=sqrt(a);
if(k*k!=a)++k;
if(b>=k)
{
printf("0\n");
continue;
}
LL t=a;
fac[0].clear(),fac[1].clear();
for(int i=1; i<=cnt&&prime[i]*prime[i]<=t; ++i) ///质因数分解
{
if(t%prime[i])
continue;
int tmp=0;
fac[0].push_back(prime[i]);
while(t%prime[i]==0)
++tmp,t/=prime[i];
fac[1].push_back(tmp);
}
if(t>1)
{
fac[0].push_back(t);
fac[1].push_back(1);
}
tot=0;
dfs(0,1);
int ans=0;
for(int i=1; i<=tot; ++i)
if(divisors[i]>=b)
++ans;
printf("%d\n",ans);
}
return 0;
}
#include <iostream>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <cstdio>
using namespace std;
typedef long long LL;
const int MAXN = 1e6+5;
bool prime[MAXN];
LL p[MAXN],k;
void isprime()
{
memset(prime, true, sizeof(prime));
prime[1] = false;
k = 0;
for(LL i=2; i<MAXN; i++)
{
if(prime[i])
{
p[k++] = i;
for(LL j=i*i; j<MAXN; j+=i)
prime[j] = false;
}
}
}
LL Solve(LL m)
{
LL ret = 1;
for(LL i=0; p[i]*p[i]<=m&&i<k; i++)
{
LL cnt = 0;
if(m%p[i] == 0)
{
while(m%p[i] == 0)
{
cnt++;
m /= p[i];
}
ret *= (cnt+1);
}
}
if(m > 1)
ret *= 2;
return ret;
}
int main()
{
isprime();
int T;
cin>>T;
for(int cas=1; cas<=T; cas++)
{
LL s,a;
cin>>s>>a;
if(a*a >= s)
printf("Case %d: 0\n",cas);
else
{
LL ret = Solve(s);
ret /= 2;
for(LL i=1; i<a; i++)
if(s % i == 0)
ret--;
printf("Case %d: %lld\n",cas,ret);
}
}
return 0;
}