使用n!近似公式算一下就可以了
注意特判
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<cmath>
using namespace std;
const long double Pi=acos(-1),e=exp(1),eps=1e-10;
long long n,k;
long double log(long double a,long double b){return log(a)/log(b);}
int main(){
freopen("3000.in","r",stdin);
freopen("3000.out","w",stdout);
while(scanf("%lld %lld",&n,&k)!=EOF){
double ans=0;
if(n<=10000){
for(int i=1;i<=n;i++)ans+=log(i);
ans/=log(k);
ans=ceil(ans+eps);
printf("%.0lf\n",ans);
}
else printf("%lld\n",(long long)(0.5*log(2*Pi*n,k)+n*log(n,k)-n*log(e,k)+1));
}
return 0;
}
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