SPOJ FACVSPOW 数论+二分

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 thang(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:3234Output:479

题意：输入a，找到满足n!>a^n 最小的n。

题解：

n*(n-1)*...*2*1>a*a*a...*a

ln(n)+ln(n-1)+...+ln(2)+ln(1)>ln(a)+ln(a)...+ln(a)

ln(a)<(ln(n)+ln(n-1)+...+ln(1))/n

预处理完二分查找

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>using namespace std;double pre[3000005];int main(){int n,t,i;double now=0;for(i=1;i<=3000000;i++){now+=log(i);pre[i]=now/i;}scanf("%d",&t);while(t--){scanf("%d",&n);int l=1,r=3000000;double d=log(n);while(l+1<r){int mid=l+r>>1;if(d<pre[mid])r=mid;else l=mid;}if(d<pre[l])printf("%d\n",l);else printf("%d\n",r);}return 0;}