int count = 0; for(int i = 0;i < 100;i++) { count = count++; } System.out.println("count = "+count);
时间: 2024-06-03 12:06:49 浏览: 12
This code will cause an infinite loop because the variable "i" in the for loop is not being incremented. Therefore, the loop condition will always be true and the loop will never end. The count variable is also not being updated in any way, so its value will remain at 0.
If we assume that the increment statement was accidentally left out of the for loop, the corrected code would be:
int count = 0;
for(int i = 0; i < 100; i++) {
count = count + 1;
}
System.out.println("count = " + count);
This code will iterate through the loop 100 times, incrementing the count variable each time. It will then print out the value of count, which should be 100.
相关问题
Scanner sc = new Scanner(System.in); int num = sc.nextInt(); for (int i = 0; i <= num; i++) { int count = 1; int tmp = i; while (count != 0) { count++; tmp = tmp / 10; } tmp = i; int sum = 0; while (tmp != 0) { sum += Math.pow(tmp % 10,count); tmp /= 10; } if(sum == i) { System.out.println(i); } }这个代码有啥问题
这段代码还是判断每个数是否为阿姆斯特朗数的代码。但是在第一个while循环中,判断条件不正确。当tmp除以10的商为0时,应该停止循环,但是判断条件是count不等于0,这是不正确的。应该改为当tmp大于0时,继续循环,当tmp等于0时,停止循环。修改后的代码如下:
```
Scanner sc = new Scanner(System.in);
int num = sc.nextInt();
for (int i = 0; i <= num; i++) {
int count = 1;
int tmp = i;
while (tmp > 0) {
count++;
tmp = tmp / 10;
}
tmp = i;
int sum = 0;
while (tmp != 0) {
sum += Math.pow(tmp % 10,count);
tmp /= 10;
}
if(sum == i) {
System.out.println(i);
}
}
```
这样修改后,就可以正确地判断每个数是否为阿姆斯特朗数了。
int n = sl.nextInt(); int count = 0; for (int i = 1; i <= n; i++) { String str = Integer.toString(i); for (int j = 0; j < str.length(); j++) { if (str.charAt(j) == '1') { count++; } } } System.out.println(count); j加个n(1≤n≤10000)
是的,您是对的,在内部循环中,j 的终止条件应该是 j <= i 的位数,而不是 j < str.length()。因为 i 的位数最多是 log10(i)+1,因此可以将内部循环的终止条件改为 j <= (int)(Math.log10(i) + 1)。修改后的代码如下:
```java
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
int n = scanner.nextInt();
int count = 0;
for (int i = 1; i <= n; i++) {
int digits = (int)(Math.log10(i) + 1);
for (int j = 0; j < digits; j++) {
int digit = i / (int)Math.pow(10, j) % 10;
if (digit == 1) {
count++;
}
}
}
System.out.println(count);
}
}
```
这个程序的思路与之前介绍的数学方法类似,但是使用了字符串转换的方式。同时,对于每个数,可以通过计算它的位数,然后依次计算每一位上的数字,来统计数字 1 出现的次数。
相关推荐
![pdf](https://img-home.csdnimg.cn/images/20210720083512.png)
![docx](https://img-home.csdnimg.cn/images/20210720083331.png)
![docx](https://img-home.csdnimg.cn/images/20210720083331.png)
![](https://csdnimg.cn/download_wenku/file_type_ask_c1.png)
![](https://csdnimg.cn/download_wenku/file_type_ask_c1.png)
![](https://csdnimg.cn/download_wenku/file_type_ask_c1.png)
![](https://csdnimg.cn/download_wenku/file_type_ask_c1.png)
![](https://csdnimg.cn/download_wenku/file_type_ask_c1.png)
![](https://csdnimg.cn/download_wenku/file_type_ask_c1.png)
![](https://csdnimg.cn/download_wenku/file_type_ask_c1.png)
![](https://csdnimg.cn/download_wenku/file_type_ask_c1.png)
![](https://csdnimg.cn/download_wenku/file_type_ask_c1.png)
![](https://csdnimg.cn/download_wenku/file_type_ask_c1.png)
![](https://csdnimg.cn/download_wenku/file_type_ask_c1.png)
![](https://csdnimg.cn/download_wenku/file_type_ask_c1.png)
![](https://csdnimg.cn/download_wenku/file_type_ask_c1.png)