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

 1 #include <iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<algorithm>
 5 #include<stdlib.h>
 6 #include<vector>
 7 #include<cmath>
 8 #include<queue>
 9 #include<set>
10 #include<map>
11 using namespace std;
12 #define N 100000
13 #define LL long long
14 #define INF 0xfffffff
15 const double eps = 1e-8;
16 const double pi = acos(-1.0);
17 const double inf = ~0u>>2;
18 int main()
19 {
20     int i,j,n;
21     //cout<<log10(10)<<" "<<log(10)<<endl;
22     while(cin>>n)
23     {
24         if(!n) break;
25         LL k = 2;
26         for(i = 1960 ; i <= n ; i+=10)
27         k*=2;
28         double d=0;
29         int ans;
30         for(i = 1; ; i++)
31         {
32             d+=log10(i*1.0)/log10(2.0);
33             if(d>k)
34             {
35                 ans = i-1;
36                 break;
37             }
38         }
39         cout<<ans<<endl;
40     }
41     return 0;
42 }
View Code

 

posted @ 2014-03-24 00:13  _雨  阅读(179)  评论(0)  编辑  收藏  举报