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 }

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