BZOJ 3000(Big Number-Stirling公式求n!近似值)

3000: Big Number

Time Limit: 2 Sec  Memory Limit: 128 MB
Submit: 220  Solved: 62
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Description

给你两个整数N和K，要求你输出N！的K进制的位数。

Input

有多组输入数据，每组输入数据各一行，每行两个数——N，K

 

Output

       每行一个数为输出结果。

 

Sample Input

2 5

2 10

10 10

100 200

Sample Output

1

1

7

69




对于100%的数据，有2≤N≤2^31， 2≤K≤200，数据组数T≤200。

HINT

Source

本题需要用到Stirling公式

∴位数=log10(n!)+1

方案1:[WA]n！不断div k

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Lson (x<<1)
#define Rson ((x<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
long long mul(long long a,long long b){return (a*b)%F;}
long long add(long long a,long long b){return (a+b)%F;}
long long sub(long long a,long long b){return (a-b+(a-b)/F*F+F)%F;}
typedef long long ll;
typedef long double ld;
ld log(ll a,int b){}
int main()
{
//freopen("bzoj3000.in","r",stdin);
//freopen(".out","w",stdout);
ll n;int k;
while (scanf("%lld%d",&n,&k)!=EOF)
{
if (n<=10000)
{
ll ans=0,p=1;
Fork(i,2,n) 
{
p*=i;
while (p>=k) {p/=k,ans++;}
cout<<i<<':'<<p<<' '<<ans<<endl;
}
if (p) ans++;
printf("%lld\n",ans);
}
}


return 0;
}

但是这样做是错的。TNT

理由是不断div k 会把原来小的部分删掉

方案2：logk(n!)=log(1)+log(2)+...+log(n)

效率为O(n)，显然也不行

于是不妨用Stirling公式求近似值

位数＝logk(n!)+1

≈ logk(sqrt(2πn)*(n/e)^n+1

= logk(sqrt(2πn))+log[(n/e)^n]+1

=1/2*logk(2πn)+nlog(n/e)+1

=0.5*logk(2πn)+nlog(n/e)+1

   =0.5*logk(2πn)+nlog(n)-nlog(e)+1

PS:pi=acos(-1.0),e=exp(1)

PS2:eps的存在是为了防止n=2,k=2这样刚好的情况出现，这个时候向上取整要多取1位

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Lson (x<<1)
#define Rson ((x<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
long long mul(long long a,long long b){return (a*b)%F;}
long long add(long long a,long long b){return (a+b)%F;}
long long sub(long long a,long long b){return (a-b+(a-b)/F*F+F)%F;}
typedef long long ll;
typedef long double ld;
ld const pi=acos(-1.0),e=exp(1),eps=1e-10;
ld log(ld a,ld b){return log(a)/log(b);}
int main()
{
//freopen("bzoj3000.in","r",stdin);
//freopen(".out","w",stdout);
ll n;int k;
cout.setf(ios::fixed);
cout.precision(0);

while (scanf("%lld%d",&n,&k)!=EOF)
{
if (n<=10000)
{
ld ans=0.0;
For(i,n) ans+=log(i);
ans/=log(k);
ans=ceil(ans+eps);
cout<<ans<<endl;
}
else 
{
cout<<ll(0.5*log(2*pi*n,k)+n*log(n,k)-n*log(e,k))+1<<endl;
}

}


return 0;
}