(HDU) 1018 big number

The question is:

Log10 (N !) = Log10 (1*2*3 *....... * n) = log10 (1) + log10 (2) + ........ + log10 (N );

If you type a table, it will be MLE and recursive. The background data will not be very smooth.

#include<cmath>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;typedef long long LL;const int maxn = 10000000;//double dp[maxn + 1];//void List(){//    dp[1] = log10(1.0);//    for(int i = 2; i < maxn; i++)//        dp[i] = dp[i - 1] + log10(1.0 * i);//    return;//}int main(){    int T;    //List();    scanf("%d",&T);    while(T--){        int n;        scanf("%d",&n);        double ret = 0;        for(int i = 1; i <= n; i++)            ret += log10(1.0 * i);        printf("%.f\n",ceil(ret));    }    return 0;}

Another method is the sterling formula.

That is, log10 (N !) = Log (N !) /Log (10) = (N * log (N)-N + 0.5 * log (2 * π * n)/log (N );

The formula is provided, which is faster.

I will not talk about it here.

(HDU) 1018 big number