Print the numbers between 30 to 3000.
Microsoft Interview Question Developer Program Engineers
看到一个题目比较有意思:
Print the numbers between 30 to 3000.
CONSTRAINT:
The numbers shouldnt contain digits either in incresing order or decreasing order.
FOLLOWING NOT ALLOWED
##123,234,345,1234,2345##increasing order,
##32,21,321,432,3210 etc##decresing order.
FOLLOWING ALLOWED:
243,27,578,2344 etc.,
Now see who ll code ths....
-
答案是:
String sortedNumbers = "123456789 9876543210";
for (int i = 31; i <= 3000; i++) {
String temp = "" + i;
if (!sortedNumbers.contains(temp)) {
System.out.println(temp);
}
}
Print the numbers between 30 to 3000.的更多相关文章
- PAT A1127 ZigZagging on a Tree (30 分)——二叉树,建树,层序遍历
Suppose that all the keys in a binary tree are distinct positive integers. A unique binary tree can ...
- 1127 ZigZagging on a Tree (30 分)
1127 ZigZagging on a Tree (30 分) Suppose that all the keys in a binary tree are distinct positive in ...
- PAT甲级 1127. ZigZagging on a Tree (30)
1127. ZigZagging on a Tree (30) 时间限制 400 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue ...
- [图的遍历&多标准] 1087. All Roads Lead to Rome (30)
1087. All Roads Lead to Rome (30) Indeed there are many different tourist routes from our city to Ro ...
- [并查集] 1107. Social Clusters (30)
1107. Social Clusters (30) When register on a social network, you are always asked to specify your h ...
- PAT A1127 ZigZagging on a Tree (30 分)
Suppose that all the keys in a binary tree are distinct positive integers. A unique binary tree can ...
- pat1087. All Roads Lead to Rome (30)
1087. All Roads Lead to Rome (30) 时间限制 200 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yu ...
- pat 甲级 1127. ZigZagging on a Tree (30)
1127. ZigZagging on a Tree (30) 时间限制 400 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue ...
- PAT-1107 Social Clusters (30 分) 并查集模板
1107 Social Clusters (30 分) When register on a social network, you are always asked to specify your ...
随机推荐
- Controlling Site Provisioning Process with a Custom Provider
http://www.cnblogs.com/frankzye/archive/2010/09/07/1820346.html http://sujoysharepoint2010.blogspot. ...
- python学习笔记26(python中__name__的使用)
在python中,每个py文件都是一个模块,也都是一个可执行文件,即包含main方法.因此,对每个py文件,可以单独运行,也可以import它给其他客户使用,这两种情况不一样. 1. 如果模块是被导入 ...
- eclipse 书签
虽然eclipse有back to和forward两个功能帮助我们阅读代码,但有时候代码一层一层看下去后,会忘了自己最初的起点. 因此想到了eclipse的书签bookmark功能. 首先,添加书签. ...
- uva 10780
曾经做过一个类似的 求n!中有多少个质因子m 这里有一个结论 k = n/m+n/(m^2)+n/(m^3)+.... int getnum(int n, int m) { int sum = 0; ...
- CodeForces 32C
额 找找规律吧 要用long long 才过. #include <cstdio> #include <algorithm> using namespace std; in ...
- "Principles of Reactive Programming" 之<Actors are Distributed> (2)
Actor Path 我们知道actor是有层级的(hierarchical),第.每个actor在它的父actor的名字空间下都有一个名字.这样就构成了一个树状的结构,就像是文件系统.每个actor ...
- C: 数组形参
知识这个东西,真是知道的越多就不知道的越多,C/C++这塘水得多深啊,哈哈.看下面3个片段:<一> 1 void fun(char a[100]) { 2 fprintf( ...
- Java 另一道构造器与构造器重载的题目
题目: 请写出以下程序的输出结果 public class ConstructorTest2 { public static void main(String[] args) { new B(&quo ...
- 深入理解Java内存模型(二)——重排序
本文属于作者原创,原文发表于InfoQ:http://www.infoq.com/cn/articles/java-memory-model-2 数据依赖性 如果两个操作访问同一个变量,且这两个操作中 ...
- UIActinSheet和UIActionSheetDelegate
UIActinSheet和UIActionSheetDelegate 这个是就那个UIActionSheet对象 一般用来选择类型或者改变界面...还有更多应用 定义如下:UIActionSheet ...