SPOJ FACVSPOW - Factorial vs Power

最新推荐文章于 2018-04-23 22:35:27 发布

WeYoungg

最新推荐文章于 2018-04-23 22:35:27 发布

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分类专栏： SPOJ 数学

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本文探讨了两个整数序列f(n)=n!与g(n)=a^n的关系，提出了一种利用斯特林公式和二分查找的方法来找到使得n! > a^n成立的最小正整数n。

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FACVSPOW - Factorial vs Power

Consider two integer sequences f(n) = n! and g(n) = an, where n is a positive integer. For any integer a > 1 the second sequence is greater than the first for a finite number of values. But starting from some integer k, f(n) is greater than g(n) for all n >= k. You are to find the least positive value of n for which f(n) > g(n), for a given positive integer a > 1.

Input

The first line of the input contains number t – the amount of tests. Then t test descriptions follow. Each test consist of a single number a.

Constraints

1 <= t <= 100000
2 <= a <= 106

Output

For each test print the least positive value of n for which f(n) > g(n).

Example

Input:
3
2
3
4

Output:
4
7
9

对于给定的a，求满足的 n! > an 最小的n

取对数，然后发现[ln(1)+ln(2)+ln(3)+ln(4)+ ……+ln(n)]/n 和n是线性关系的，所以可用二分来求满足＞n*ln(a)的最小n

哦对学到了一个很有趣的斯特林公式

ln(n!) = n * ln(n) - n + 0.5*(ln(PI * 2 * n))

$n! \approx \sqrt{2\pi n}\, \left(\frac{n}{e}\right)^{n}.$

#include<cmath>
#include<algorithm>
#include<cstring>
#include<string>
#include<set>
#include<map>
#include<time.h>
#include<cstdio>
#include<vector>
#include<stack>
#include<queue>
#include<iostream>
using namespace std;
#define  LONG long long
const int   INF=0x3f3f3f3f;
const int MOD=1e9+7;
const double PI=acos(-1.0);
#define clrI(x) memset(x,-1,sizeof(x))
#define clr0(x) memset(x,0,sizeof x)
#define clr1(x) memset(x,INF,sizeof x)
#define clr2(x) memset(x,-INF,sizeof x)
#define EPS 1e-10
bool check(LONG n ,double a )
{
    double x = (double )n;
    double sum = x * log(x) - x + 0.5*(log(PI * 2 * x));
    if(sum > x * log(a))
        return 1;
    else return 0;
}
int main()
{
    int T;
    cin>>T;
    while(T--)
    {
        double  a ;
        cin>>a ;
        LONG l = 1 , r=10*a;
        LONG mid  ;
        while(l < r)
        {
            mid = (l  + r ) / 2;
            if(check(mid,a))
            {
                r = mid ;
            }
            else
                l = mid + 1;
         //   cout<<mid<<endl;
        }
        cout<<l<<endl;
    }
}