POJ 1423 Big Number

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这个超时

#include<iostream>#include<math.h>#include<fstream>#include <time.h>using namespace std;#define Len 5500000double s[Len+1];int main(){    s[0]=0;    clock_t start = clock();    for(int i=1;i<=Len;++i){        s[i]=log10(i)+s[i-1];    }    clock_t endt = clock();    cout<<"t:"<<(double)(endt -start)/CLOCKS_PER_SEC<<endl;    //ifstream cin("1.txt");    int t,n;    cin>>t;    while(t--)    {        cin>>n;        if(n<=Len)            cout<<(int)s[n]+1<<endl;        else{            double sum=s[Len];            for(int i=Len+1;i<=n;++i){               sum+=log10(i);            }            cout<<(int)sum+1<<endl;        }    }    return 0;}

这个应用公式

#include <iostream>#include <math.h>#include <fstream>using namespace std;int n;const double e = 2.7182818284590452354, pi = 3.141592653589793239;double f( int a ){    return log10( sqrt( 2 * pi * a ) ) + a * log10( a / e );}int main(){    int cas, ans;    double i, s;    cin>>cas;    while( cas-- )    {        cin>>n;        if( n < 100000 )        {            for( s=0, i=1; i<=n; i++ )                s += log10( i );        }        else s = f( n );        ans = (int)s;        if( ans <= s )            ans++;        cout<<ans<<endl;    }    return 0;}

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