ZOJ 2277 (数论)

标签：iostream   acm   zoj   math   

Background

It is dark at night.
It is silence at night.
It is she in the dark.
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;
}

ZOJ 2277 (数论)

标签：iostream   acm   zoj   math   

原文地址：http://blog.csdn.net/u012844301/article/details/47168847