下面的代码段执行之后count的值是什么( ) int count = 1; for(int i = 1; i <= 5; i++) { count+= i; } System.out.println(count); int count = 1; for(int i = 1; i <= 5; i++) { count+= i; } System.out.println(count);
举一反三
- 下面的代码段执行之后count的值是什么( ) int count = 1; for (int i = 1; i <= 5; i++) { count += i; } System.out.println(count);
- 下列程序的运行结果是（）。 public class Test public static void main ( String [ ] args ) int count = 0 for( int i = 1 i < 5 i = 2) for( int j = 1 j< = 10 j = 3) count System .out .print (count ) _
- 如果希望实现从1到10的累加,并得出结果;int count=0;_______{count +=i;}应该在空白处添加下面哪段代码?( ) A: for (int i=0; i<;10;i++) B: for (int i=0; i<;=9;i++) C: for (int i=1; i<;=10;i++) D: for (int i=1; i<;10;i++)
- 写出下列程序的输出结果 public class Test { public static void main(String[] args) { Count myCount = new Count(); int times = 0; for(int i=0;i<100;i++) increment(myCount , times); System.out.println(“count is” + myCount.count); System.out.println(“time is”+ times); } public static void increment(Count c , int times) { c.count++; times++; } } class Count { public int count; Count(int c) { count =c; } Count() { count =1; } }
- 下列程序中sumFun()的算法时间复杂度为 ( ) 。int sumFun(int n) { int count = 0; for (int i = 1; i < n; i = i * 2) for (int j = i; j > 0; j = j - 1) count = count + i + j; return count; } A: O(n) B: O(nlogn) C: O(n^2) D: O(logn logn)