poj2661Factstone Benchmark

链接

利用log函数来求解 n!<=2^k k会达到400+W 暴力就不要想了，不过可以利用log函数来做

log2(n!) = log2(1)+log2(2)+..log2(n)<=k

 #include <iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<stdlib.h>
 #include<vector>
 #include<cmath>
 #include<queue>
 #include<set>
 #include<map>
 using namespace std;
 #define N 100000
 #define LL long long
 #define INF 0xfffffff
 const double eps = 1e-;
 const double pi = acos(-1.0);
 const double inf = ~0u>>;
 int main()
 {
     int i,j,n;
     //cout<<log10(10)<<" "<<log(10)<<endl;
     while(cin>>n)
     {
         if(!n) break;
         LL k = ;
         for(i =  ; i <= n ; i+=)
         k*=;
         double d=;
         int ans;
         for(i = ; ; i++)
         {
             d+=log10(i*1.0)/log10(2.0);
             if(d>k)
             {
                 ans = i-;
                 break;
             }
         }
         cout<<ans<<endl;
     }
     return ;
 }