3000: Big Number
Time Limit: 2 Sec Memory Limit: 128 MB
Submit: 220 Solved: 62
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Description
給你兩個整數N和K，要求你輸出N！的K進制的位數。
Input
有多組輸入資料，每組輸入資料各一行，每行兩個數——N，K
Output
每行一個數為輸出結果。
Sample Input
2 5
2 10
10 10
100 200
Sample Output
1
1
7
69
對于100%的資料，有2≤N≤2^31， 2≤K≤200，資料組數T≤200。
HINT
Source
本題需要用到Stirling公式
∴位數=log10(n!)+1
方案1:[WA]n！不斷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;
}
但是這樣做是錯的。TNT
理由是不斷div k 會把原來小的部分删掉
方案2：logk(n!)=log(1)+log(2)+...+log(n)
效率為O(n)，顯然也不行
于是不妨用Stirling公式求近似值
位數＝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的存在是為了防止n=2,k=2這樣剛好的情況出現，這個時候向上取整要多取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;
}