POJ 1552-Doubles
来源:互联网 发布:mac搜索不到蓝牙音箱 编辑:程序博客网 时间:2024/06/06 21:18
Description
As part of an arithmetic competency program, your students will be given randomly generated lists of from 2 to 15 unique positive integers and asked to determine how many items in each list are twice some other item in the same list. You will need a program to help you with the grading. This program should be able to scan the lists and output the correct answer for each one. For example, given the list
your program should answer 3, as 2 is twice 1, 4 is twice 2, and 18 is twice 9.
1 4 3 2 9 7 18 22
your program should answer 3, as 2 is twice 1, 4 is twice 2, and 18 is twice 9.
Input
The input will consist of one or more lists of numbers. There will be one list of numbers per line. Each list will contain from 2 to 15 unique positive integers. No integer will be larger than 99. Each line will be terminated with the integer 0, which is not considered part of the list. A line with the single number -1 will mark the end of the file. The example input below shows 3 separate lists. Some lists may not contain any doubles.
Output
The output will consist of one line per input list, containing a count of the items that are double some other item.
Sample Input
1 4 3 2 9 7 18 22 02 4 8 10 07 5 11 13 1 3 0-1
Sample Output
320
Source
Mid-Central USA 2003
题目大意:一串数,以0结尾,找出一个数是另一个数的两倍的有几组数。
思路:水。
代码:
import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
boolean flag = true;
int a[] = new int[16];
while (scan.hasNext() && flag) {
int k = 0;
int ans = 0;
a = new int[16];
while (true) {
int b = scan.nextInt();
if (b == 0)
break;
if (b == -1) {
flag = false;
break;
}
a[k] = b;
k++;
}
if(flag){
for (int i = 0; i <k; i++) {
for (int j = 0; j <k; j++) {
if (a[j] == a[i] * 2) {
ans += 1;
}
}
}
System.out.println(ans);}
}
Scanner scan = new Scanner(System.in);
boolean flag = true;
int a[] = new int[16];
while (scan.hasNext() && flag) {
int k = 0;
int ans = 0;
a = new int[16];
while (true) {
int b = scan.nextInt();
if (b == 0)
break;
if (b == -1) {
flag = false;
break;
}
a[k] = b;
k++;
}
if(flag){
for (int i = 0; i <k; i++) {
for (int j = 0; j <k; j++) {
if (a[j] == a[i] * 2) {
ans += 1;
}
}
}
System.out.println(ans);}
}
}
}
0 0
- POJ 1552 Doubles
- poj 1552 Doubles 水
- poj 1552 Doubles
- 【POJ-1552】Doubles
- Poj 1552 Doubles(水题)
- POJ 1552 Doubles
- poj 1552 Doubles
- POJ 1552 - Doubles
- Doubles - POJ 1552 水题
- POJ 1552 Doubles
- POJ 1552-Doubles
- POJ 1552 Doubles
- POJ 1552 Doubles
- POJ 1552 Doubles
- POJ 1552 Doubles
- POJ 1552 Doubles 水
- Doubles POJ 1552
- POJ 1552 Doubles G++
- 收集路由器的默认密码
- 魔界/指环王三部曲(加长版)在线观看免费bt下载
- jQuery -> 获取元素的各种过滤器(filter)
- String.Format,DateTime日期时间格式化集锦
- Java配置环境变量
- POJ 1552-Doubles
- Unity3d 镜面折射 vertex and frag Shader源码
- LR的两种推导方式
- centos 6.5安装的UEFI-GPT回退为MBR引导
- 日积月累:LinearLayout的andrid:layout_weight属性的使用详解
- Tomcat+DLink路由器+花生壳搭建外网访问内网工程
- Redis新的存储模式diskstore
- OpenCV访问Mat对象中数据时发生异常---Mat中的数据访问
- Erlang进程堆垃圾回收机制