ZOJ 2277 (数论)

The Gate to Freedom

It is dark at night.

It is she in the silence.

Then a light appeared. A huge gate came into our sights, called

The Gate to Freedom

Problem

There‘re some words on the gate: "This gate will lead you to freedom. First, you have to open it. I have a problem for you to solve, if you answer it correctly, the gate will open!"

"Tell me, young boy, what is the leftmost digit of N^N?"

Input

This problem contains multiple test cases.

Each test case contains an integer N (N<=1,000,000,000).

Output

For each test case, output the leftmost digit of N^N.

Sample Input

3

4

Sample Output

2

2

Contest: A Great Beloved and My Gate to Freedom

There is a cx, there is a love_cx.

题意：

给定一个数字n , 问n的n次方的首位是多少.

解题思路：

直接暴力算n的n次方明显是不行的，用C写肯定会爆long long ，想一边暴力计算一边取模的同学可以打表看看，数字小的时候不会有误差，但是，数字大的时候会有1的误差。

而用JAVA的大数来进行计算则会超时。

那么只能从数学方面入手了。

n^n = k;

写成科学计数法后 k = a * 10^b

即  n^n = a * 10^b

两边取对数后

lg(n^n) = lg(a * 10^b)

即 n*lg(n) = lg(a) + b

那么b为多少呢?

因为 10^(n*lg(n)) = n^n

可知 [n*lg(n)] （[n*lg(n)] = n*lg(n) 向下取整）即等于 n^n 的位数  即 b

因此 n^n 的数字部分为 10^(n*lg(n) - [n*lg(n)])

我们只需要取第一位即可

#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <queue>
#include <map>
#include <set>
#include <stack>
#include <vector>
#include <sstream>
#define PI acos(-1.0)
#define eps 1e-8
const int  inf =  (1<<30) - 10;
using namespace std;

int main(){
    //freopen("input.txt","r",stdin);
    double n;
    while(cin>>n){
        double ans = ((log10(n))*n);
        ans -= floor(ans);
        ans = floor(pow(10.0,ans) + eps);
        printf("%.6f\n",ans);
    }
    return 0;
}

如有BUG，请大家务必指出，不胜感激~

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