国产探花免费观看_亚洲丰满少妇自慰呻吟_97日韩有码在线_资源在线日韩欧美_一区二区精品毛片,辰东完美世界有声小说,欢乐颂第一季,yy玄幻小说排行榜完本

首頁 > 學(xué)院 > 開發(fā)設(shè)計(jì) > 正文

hdu--1060--Leftmost Digit

2019-11-11 03:57:57
字體:
供稿:網(wǎng)友

Leftmost Digit Time Limit: 2000/1000 MS (java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 17741 Accepted Submission(s): 6854

PRoblem Description

Given a positive integer N, you should output the leftmost digit of N^N.

Input

The input contains several test cases. The first line of the input is a single integer T which is the number of test cases. T test cases follow. Each test case contains a single positive integer N(1<=N<=1,000,000,000).

Output

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

Sample Input

2 3 4

Sample Output

2 2

Hint

In the first case, 3 * 3 * 3 = 27, so the leftmost digit is 2. In the second case, 4 * 4 * 4 * 4 = 256, so the leftmost digit is 2.


需要用到科學(xué)記數(shù)法和對(duì)數(shù)運(yùn)算的知識(shí)。 我們把num*num的值記作:num*num=a*10^n,其中1

附上ac代碼:

#include<cstdio>#include<cmath>#include<cstring>#include<iostream>#include<algorithm>using namespace std;typedef long long ll;int main(){ int t; scanf ("%d",&t); while (t--) { ll n; int i,j; double x; scanf ("%lld",&n); x = n*log10(n*(1.0)); x -= (ll)x; int a = pow(10.0,x); printf ("%d/n",a); } return 0;}
發(fā)表評(píng)論 共有條評(píng)論
用戶名: 密碼:
驗(yàn)證碼: 匿名發(fā)表
主站蜘蛛池模板: 岚皋县| 莱州市| 保山市| 云龙县| 晋州市| 峨边| 苍溪县| 灵台县| 武夷山市| 二连浩特市| 益阳市| 昂仁县| 武冈市| 灵台县| 兴安县| 聂拉木县| 隆德县| 安阳市| 龙南县| 龙山县| 西青区| 上杭县| 石棉县| 庄河市| 泸州市| 宁城县| 平昌县| 三原县| 双桥区| 横山县| 天水市| 绥滨县| 威海市| 常宁市| 南陵县| 黄大仙区| 攀枝花市| 土默特右旗| 赫章县| 麟游县| 曲水县|