如何移动PostgreSQL的什么叫大数据分析目录移到新位置上的Ubuntu 16.04

来源：蜘蛛抓取(WebSpider) 时间：2017-11-08 07:12 标签： 什么叫大数据分析
求解 这段代码用java如何编写。_java吧_百度贴吧
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求解 这段代码用java如何编写。收藏
1007. To and Fro
DescriptionMo and Larry have devised a way of encrypting messages. They first decide secretly on the number of columns and write the message (letters only) down the columns, padding with extra random letters so as to make a rectangular array of letters. For example, if the message is "There's no place like home on a snowy night" and there are five columns, Mo would write down t o i o y h p k n n e l e a i r a h s g e c o n h s e m o t n l e w x Note that Mo includes only letters and writes them all in lower case. In this example, Mo used the character `x' to pad the message out to make a rectangle, although he could have used any letter. Mo then sends the message to Larry by writing the letters in each row, alternating left-to-right and right-to-left. So, the above would be encrypted as toioynnkpheleaigshareconhtomesnlewx Your job is to recover for Larry the original message (along with any extra padding letters) from the encrypted one. InputThere will be multiple input sets. Input for each set will consist of two lines. The first line will contain an integer in the range 2 . ..20 indicating the number of columns used. The next line is a string of up to 200 lower case letters. The last input set is followed by a line containing a single 0, indicating end of input. OutputEach input set should generate one line of output, giving the original plaintext message, with no spaces. Sample Input5toioynnkpheleaigshareconhtomesnlewx3ttyohhieneesiaabss0Sample Outputtheresnoplacelikehomeonasnowynightxthisistheeasyoneab 代码#include &iostream&#include &cstring&int main(){
int i, j, k,
int numberOfC
cin && numberOfC
char useless[1];
cin.getline(useless,1);
char message[201];
while(numberOfColumn != 0)
cin.getline(message,201);
int numberOfRow = strlen(message) / numberOfC
for(j=1; message[j * numberOfColumn] &= 'a' && message[j * numberOfColumn] &= 'z'; j++)
if (j % 2 == 0)
for (k = 0;k & numberOfColumn / 2; k++)
temp = message[(j + 1) * numberOfColumn - 1 - k];
message[(j + 1) * numberOfColumn - 1 - k] = message[j * numberOfColumn + k];
message[j * numberOfColumn + k] =
for(i = 0; i & numberOfC i++)
for(j = 0; j & numberOfR j++)
cout && message[numberOfColumn * j + i];
cin && numberOfC
cin.getline(useless,1);
return 0;}
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啊啊啊啊啊 求帮助 。刚开始学java 不知道从哪里下手。
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为兴趣而生，贴吧更懂你。或【求助】java程序中有一段代码编译总是提示warning_软件吧_百度贴吧
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【求助】java程序中有一段代码编译总是提示warning收藏
public String[] getItemsProperties() {
String itemsProperties[] = new String[getNumberOfItems()];
int i = 0;
for(（Iterator） iterator1 = PCBItems.iterator(); iterator1.hasNext();)
ProcessPCB stu_Item = (ProcessPCB)iterator1.next();
itemsProperties[i++] = stu_Item.toString();
return itemsP }总在Iterator（红括号括起来的地方）那里提示Iterator is a raw type. References to generic type Iterator&E& should be parameterized求解答
达内软件开发培训,&5+4&革命式软件测试培训,8大企业级实战项目,名师实操.高端软件测试课程覆盖Selemium,Jmeter,QTP,QC等高精软件测试工具,高薪入名企
Iterator 改为 Iterator&ProcessPCB &
就不会有警告了，上面的警告是泛型警告
1.用JDK1.5以前的版本2.或着用@SuppressWarnings("unchecked")3.如果是在Eclipse下开发的话，还可以更改Eclipse的编译属性，在 windows-设定-java-compile-错误和警告 里面有不检查泛型 这个选项。改成无视就行了Eclipse--&首选项---&JAVA--&Compiler--&Errors/Warnings--&Generic types--&下面三个全部改成Ignore就OK了搬来的，自己实践操作下哈
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