int count = 0; for(int i = 0;i < 100;i++) { count = count++; } System.out.println("count = "+count);
时间: 2024-06-04 17:11:18 浏览: 15
There is a syntax error in this code. The increment statement in the for loop is missing. It should be:
int count = 0;
for(int i = 0; i < 100; i++) {
count = count + 1;
}
System.out.println("count = " + count);
This code initializes a variable called "count" to 0, then loops through 100 iterations, incrementing the count variable by 1 each time. Finally, it prints out the value of the count variable. The output should be "count = 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。