[英]number pattern programs in java
How to print the triangle below: 如何打印下面的三角形:
2 3 5 8 3 8
4 6 9 4 9
7 1 5 1
2 6 2
7 3
4
First you need to start with number 2 and add one to the next one vertically 首先,您需要从数字2开始,然后垂直添加一个到下一个
My code: 我的代码:
int d = 2, n = 6;
for (int line=1; line <= n; line++ ) {
for (int j = 2; j <= line; j++) {
System.out.print(" ");
}
for (int k = line; k <= n; k++) {
System.out.print(d + " ");
d = d + k;
if (d > 9) {
d = d - 9;
}
}
System.out.println();
}
Result: 结果:
2 3 5 8 3 8
5 7 1 5 1
7 1 5 1
7 2 7
4 9
6
The pattern is that the value of d
has to be calculated initially on every new line based on the value of d
in the first instance of the previous line . 所述图案是,该值
d
必须被基于的值的每个新行初始计算 d
在上一行的第一个实例 。 That is the part that's missed here. 这是在这里错过的部分。 You can do that by having a temp variable store the initial value of
d
on every line and print based on that. 你可以通过让temp变量在每一行存储
d
的初始值并根据它进行打印来实现。 I have used a variable tempD
here, which can help print the pattern that you require. 我在这里使用了一个变量
tempD
,它可以帮助打印你需要的模式。
int d = 2, n = 6;
int tempD = d - 1;
for (int line = 1; line <= n; line++) {
tempD = tempD + line;
if (tempD > 9) {
tempD = tempD - 9;
}
d = tempD;
for (int j = 2; j <= line; j++) {
System.out.print(" ");
}
for (int k = line; k <= n; k++) {
System.out.print(d + " ");
d = d + k;
if (d > 9) {
d = d - 9;
}
}
System.out.println();
}
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