Java 如何根据计数生成模式?
如果需要,如何生成如下模式 输入号码:2 输入行数:5Java 如何根据计数生成模式?,java,design-patterns,Java,Design Patterns,如果需要,如何生成如下模式 输入号码:2 输入行数:5 2 33 444 5555 66666 publicstaticvoidmain(字符串[]args) { System.out.println(“输入一个数字(0-9):”; 扫描仪sc1=新扫描仪(System.in); int inputNo=sc1.nextInt(); System.out.println(“输入行数:”); 扫描仪sc2=新扫描仪(System.in); int inputNoLine=sc2.nextInt(
2
33
444
5555
66666
publicstaticvoidmain(字符串[]args)
{
System.out.println(“输入一个数字(0-9):”;
扫描仪sc1=新扫描仪(System.in);
int inputNo=sc1.nextInt();
System.out.println(“输入行数:”);
扫描仪sc2=新扫描仪(System.in);
int inputNoLine=sc2.nextInt();
用于(i=输入编号;iinputNoLine)
{
打破
}
系统输出打印LN(i);
}
}
我只能像上面那样编码。有什么建议吗?
谢谢。这应该行得通
public static void main(String[] args)
{
System.out.println("Input a number(0-9):");
Scanner sc1 = new Scanner(System.in);
int inputNo = sc1.nextInt();
System.out.println("Input number of lines:");
Scanner sc2 = new Scanner(System.in);
int inputNoLine = sc2.nextInt();
int count = 1;
for (int i=0; i < inputNoLine; i++)
{
for(int j = 0; j < count; j++)
{
System.out.print(inputNo);
}
count++;
inputNo++;
System.out.println();
}
}
publicstaticvoidmain(字符串[]args)
{
System.out.println(“输入一个数字(0-9):”;
扫描仪sc1=新扫描仪(System.in);
int inputNo=sc1.nextInt();
System.out.println(“输入行数:”);
扫描仪sc2=新扫描仪(System.in);
int inputNoLine=sc2.nextInt();
整数计数=1;
for(int i=0;i
试试这个,我想这就是你需要的
public static void main(String[] args) {
System.out.println("Input a number(0-9):");
Scanner sc1 = new Scanner(System.in);
int inputNo = sc1.nextInt();
System.out.println("Input number of lines:");
Scanner sc2 = new Scanner(System.in);
int inputNoLine = sc2.nextInt();
for(int line=1 ; line <=inputNoLine; line++) {
for(int count=0; count<line; count++) {
System.out.print(inputNo);
}
System.out.println();
inputNo++;
}
}
publicstaticvoidmain(字符串[]args){
System.out.println(“输入一个数字(0-9):”;
扫描仪sc1=新扫描仪(System.in);
int inputNo=sc1.nextInt();
System.out.println(“输入行数:”);
扫描仪sc2=新扫描仪(System.in);
int inputNoLine=sc2.nextInt();
对于(int line=1;line,您可以这样做:
public static void main(String args[]) throws IOException
{
System.out.println("Input a number(0-9):");
Scanner sc1 = new Scanner(System.in);
int inputNo = sc1.nextInt();
System.out.println("Input number of lines:");
Scanner sc2 = new Scanner(System.in);
int inputNoLine = sc2.nextInt();
for (int i = 0; i < inputNoLine; ++i,++inputNo) {
for (int j = 0; j < i + 1; ++j) {
System.out.print(inputNo);
}
System.out.println();
}
}
publicstaticvoidmain(字符串args[])引发IOException
{
System.out.println(“输入一个数字(0-9):”;
扫描仪sc1=新扫描仪(System.in);
int inputNo=sc1.nextInt();
System.out.println(“输入行数:”);
扫描仪sc2=新扫描仪(System.in);
int inputNoLine=sc2.nextInt();
对于(int i=0;i
您需要此代码:
public static void main(String[] args) throws ParseException,
UnsupportedEncodingException, ClassNotFoundException {
System.out.println("Input a number(0-9):");
Scanner sc1 = new Scanner(System.in);
int inputNo = sc1.nextInt();
System.out.println("Input number of lines:");
Scanner sc2 = new Scanner(System.in);
int inputNoLine = sc2.nextInt();
for (int i = 0; i < inputNoLine; i++,inputNo++) {
for (int j = 0; j < inputNo; j++) {
System.out.print(inputNo);
}
System.out.println();
}
}
publicstaticvoidmain(字符串[]args)抛出异常,
UnsupportedEncodingException,ClassNotFoundException{
System.out.println(“输入一个数字(0-9):”;
扫描仪sc1=新扫描仪(System.in);
int inputNo=sc1.nextInt();
System.out.println(“输入行数:”);
扫描仪sc2=新扫描仪(System.in);
int inputNoLine=sc2.nextInt();
for(int i=0;i
假设输入3和输入5的输出为:
试试这个:
int times = 1;
for (int i=inputNo;i<=inputNoLine;i++){
for(int j=1;j<=times;j++) {
System.out.print(i);
}
times++;
System.out.println();
}
int次=1;
对于(int i=inputNo;模式是数字应该在第一行显示1次,在第二行显示2次,在第三行显示3次..依此类推?@user1504015如果用户输入inputNo作为3,inputNoLine作为3,模式是什么5@Bingo图案将变为34455666677777@user1504015哦,是的,然后检查我的回答有一些错误…我用工作代码。那么用户输入呢?对于您的回复,thnx。但是如果inputNo 3和inputNoLine,您的代码将生成3445555@user1504015似乎您错过了条件inti=inputNo;i@user1504015在条件部分,它的inputNoLine
notinputNo
这就是为什么只得到3行
public static void main(String args[]) throws IOException
{
System.out.println("Input a number(0-9):");
Scanner sc1 = new Scanner(System.in);
int inputNo = sc1.nextInt();
System.out.println("Input number of lines:");
Scanner sc2 = new Scanner(System.in);
int inputNoLine = sc2.nextInt();
for (int i = 0; i < inputNoLine; ++i,++inputNo) {
for (int j = 0; j < i + 1; ++j) {
System.out.print(inputNo);
}
System.out.println();
}
}
public static void main(String[] args) throws ParseException,
UnsupportedEncodingException, ClassNotFoundException {
System.out.println("Input a number(0-9):");
Scanner sc1 = new Scanner(System.in);
int inputNo = sc1.nextInt();
System.out.println("Input number of lines:");
Scanner sc2 = new Scanner(System.in);
int inputNoLine = sc2.nextInt();
for (int i = 0; i < inputNoLine; i++,inputNo++) {
for (int j = 0; j < inputNo; j++) {
System.out.print(inputNo);
}
System.out.println();
}
}
3
44
555
6666
77777
int times = 1;
for (int i=inputNo;i<=inputNoLine;i++){
for(int j=1;j<=times;j++) {
System.out.print(i);
}
times++;
System.out.println();
}