SPOJ FACVSPOW - Factorial vs Power 二分

题目链接：http://www.spoj.com/problems/FACVSPOW/

题目大意：给定底数a，问什么时候n! 超过 aⁿ. 1 <= t <= 100000    2 <= a <= 106

解题思路：等价于计算 使1*2*3*4*5*...*k / a^k  > 1 的 k。直接计算必然溢出，上下同取log，然后就不怕溢出了。预处理log前N项和，二分check找。注意使用double。

代码：

int a;
double sum[maxn];

int check(){
    int l = 1, r = 5000000;
    while(l < r){
        int mid = (l + r) >> 1;
        double up = sum[mid], dn = mid * log(a);
        if(up >= dn) r = mid;
        else l = mid + 1;
    }
    return l;
}
void dowork(){
    sum[0] = 0;
    for(int i = 1; i < 5e6; i++) sum[i] = sum[i - 1] + log(i);
}
int main(){
    dowork();
    int t;
    scanf("%d", &t);
    while(t--){
        scanf("%d", &a);
        int ans = check();
        printf("%d\n", ans);
    }
}

题目：

FACVSPOW - Factorial vs Power

#math #binary-search

 

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

 Submit solution!