Beyond The Sky

http://acm.hdu.edu.cn/game/entry/problem/show.php?chapterid=1&sectionid=1&problemid=3004import java.util.Scanner;public class Main { public static void main(String[] args) { int n; Scanner in

分析：水题import java.util.Arrays;import java.util.Scanner;public class Main { public static void main(String[] args) { Scanner in=new Scanner(System.in); int n = in.nextInt(); while(0 != n){

分析：水题。 但错了一一次，

今年暑假不ACTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 2456 Accepted Submission(s): 1348 Problem Description“

Saving HDUTime Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2343 Accepted Submission(s): 1046Problem Description

shǎ崽 OrOrOrOrzTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3942 Accepted Submission(s): 1224 Problem Descripti

分析：不用打表发，直接判断是不是素数。注意Math.sqrt(x).

分析：在做题的时候用到了需要对int

import java.util.Arrays;import java.util.Scanner;public class Main { public static void main(String[] args) { Scanner in=new Scanner(System.in); while(in.hasNext()){ int a = in.nextInt();

import java.util.Arrays;import java.util.Scanner;public class Main { public static int gcd(int m, int n) { if (m < n) { int a = m; m = n; n = a; } if (m % n == 0) { return n; }

import java.util.Scanner;public class Main { public static void main(String[] args) { int n; Scanner in = new Scanner(System.in); n = in.nextInt(); in.nextLine(); while (n-->0) { // Sys

import java.util.Scanner;public class Main { public static void main(String[] args) { int a, b; Scanner in = new Scanner(System.in); while (in.hasNext()) { String str = in.next(); a = I

import java.util.Scanner;public class Main { public static void main(String[] args) { int n,a,b; int t; Scanner in = new Scanner(System.in); n = in.nextInt(); while( n-- > 0){ int

分析：这道题目和上道题目非常像，只不过

Vowel CountingTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2460 Accepted Submission(s): 1354 Problem Descripti

http://acm.hdu.edu.cn/game/entry/problem/show.php?chapterid=1§ionid=1&problemid=3008import java.util.Scanner;public class Main { public static void main(String[] args) { int N,n,s; Scann

http://acm.hdu.edu.cn/game/entry/problem/show.php?chapterid=1&sectionid=1&problemid=3007

http://acm.hdu.edu.cn/game/entry/problem/show.php?chapterid=1&sectionid=1&problemid=3005import java.util.Scanner;public class Main { public static void main(String[] args) { int n,N; Scanner

http://acm.hdu.edu.cn/game/entry/problem/show.php?chapterid=1&sectionid=1&problemid=3006

import java.util.Scanner;public class Main { public static int pow(int x, int n) { if (x == 1) return 1; if (n == 1) return x; if(n==0 ) return 1; else if (n % 2 == 1){ int s = pow

import java.util.Scanner;public class Main { public static boolean isPrime(int x) { int flag=0; int t = (int) Math.sqrt(x); for (int i = 2; i <= t; i++) { if (x % i == 0){ flag=1;

http://acm.hdu.edu.cn/game/entry/problem/show.php?chapterid=2&sectionid=1&problemid=1

分析：这个题算法上没有什么难度，主要在于考察基本字符串，基本控制的运用。在

分析： 题目没有太大的难度，使用Ha

QuicksumTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2437 Accepted Submission(s): 1535Problem Description

下面的代码时间超时，因为使用了对Integer排序：import java.util.Arrays;import java.util.Comparator;import java.util.Iterator;import java.util.Map;import java.util.Scanner;import java.util.Set;import java.util.Tree

开门人和关门人Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 2659 Accepted Submission(s): 1250 Problem Description每

第一个是一个正确的算法，但是超时。import java.util.Scanner;public class Main { public static int N=1000000; public static void main(String[] args) { int [] a=new int[N]; int n; int ans=-2; Scanner in = n