print('{0}*{1}={2}'.format(i,j,i*j),end=' '
时间: 2024-04-28 11:01:48 浏览: 28
这段代码缺少一个右括号,应该是这样的:
```
print('{0}*{1}={2}'.format(i,j,i*j),end=' ')
```
该代码使用了字符串的格式化方法 `format()`,通过传入不同的参数,可以将字符串中的占位符 `{}` 替换为对应的值。其中,`{0}` 表示第一个参数,`{1}` 表示第二个参数,以此类推。在这个例子中,`{0}` 被替换为 `i`,`{1}` 被替换为 `j`,`{2}` 被替换为 `i*j`。`end=' '` 则表示输出后不换行,以空格结尾。
相关问题
python打印4、打印如下所示九九乘法表。 9*9=81 9*8=72 9*7=63 9*6=54 9*5=45 9*4=36 9*3=27 9*2=18 9*1=1 8*8=64 8*7=56 8*6=48 8*5=40 8*4=32 8*3=24 8*2=16 8*1=8 ...... ...... 1*1=1(10分)
好的,下面是Python代码:
```python
for i in range(9, 0, -1):
for j in range(i, 0, -1):
print("{}*{}={}".format(i, j, i*j), end=" ")
print()
```
输出结果为:
```
9*9=81 9*8=72 9*7=63 9*6=54 9*5=45 9*4=36 9*3=27 9*2=18 9*1=9
8*8=64 8*7=56 8*6=48 8*5=40 8*4=32 8*3=24 8*2=16 8*1=8
7*7=49 7*6=42 7*5=35 7*4=28 7*3=21 7*2=14 7*1=7
6*6=36 6*5=30 6*4=24 6*3=18 6*2=12 6*1=6
5*5=25 5*4=20 5*3=15 5*2=10 5*1=5
4*4=16 4*3=12 4*2=8 4*1=4
3*3=9 3*2=6 3*1=3
2*2=4 2*1=2
1*1=1
```
设计算法,能打印 n *n的乘法表(九九乘法表仅是其中一种),格式如下 1*1=1 1*2=2 .。。。 2*2=4 .。。 給出算法复杂度,并编程实现,主函数中输入数据,打印输出。
1)算法设计:可以使用两个循环来实现乘法表的输出,外层循环控制行数,内层循环控制列数。具体算法步骤如下:
- 对于每一行i,从1到n输出i*1,i*2,...,i*n
- 每行输出结束后换行
2)算法复杂度:该算法的时间复杂度为O(n^2),因为需要输出n^2个元素。
3)编程实现:
```python
def multiplication_table(n):
for i in range(1, n+1):
for j in range(1, n+1):
print("{}*{}={}".format(i, j, i*j), end="\t")
print()
if __name__ == "__main__":
n = int(input("请输入一个整数n:"))
multiplication_table(n)
```
在主函数中,首先读取键盘输入的整数n,然后调用multiplication_table函数进行乘法表的输出,最后打印输出。
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