以下代码全部在myeclipse里运行过了:
第一题:
package te;
import java.util.Scanner;
public class sum {
/**
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
int sum=0;
System.out.println("请输入n的值:");
int n;
Scanner sc=new Scanner(System.in);
n=sc.nextInt();
for(int i=1,j=-1; i<=2*n-1; i+=2)
{
j*=-1;
sum+=(2*n-1)*j;
}
System.out.println("1-3+5-7+9-11+…+(2n-1)="+sum);
}
}
第二题:
package te;
import java.util.Scanner;
public class Ecx {
/**
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
System.out.println("请输入n的值:");
int n;
Scanner sc=new Scanner(System.in);
n=sc.nextInt();
int[] array=new int[n];
for(int i=0;i
}
for(int i=0;i
}
int max=0,min=0;
for(int i=0;i
if(min>array[i])//求最小值
min=array[i];
}
System.out.println("Max="+max);//输出数组的最大值
System.out.println("Min="+min);//输出数组的最小值
}
}
什么语言啊?
pravate Sub Form_Click()
dim a , i as integer
n=Text1
a=0
For i=1 To n
a=a+(2*i -1)*(-1)^(i+1)
Nexti
print a
End sub
第一题的答案 呵呵 应该对
pravate Sub Form_Click()
dim a() as integer
max=0 : min=1000
n = inputbox("请输入n的值")
redim a(1 to n, 1 to n)
for i = 1 to n
for j = 1 to n
a(i,j) = int(Rnd*60-10+1+10)
print a(i,j)
next j
print
next i
for i = 1 to n
for j = 1 to n
if a(i,j)>max then max=a(i,j)
if a(i,j)
next i
print max , min
end sub
第二题的答案
上面的都是用的vb 解答 的
JAVA语言答第一题:
int calc(int n)
{
int res = 0;
int coef = 1;
for(int i=1; i<=2*n-1; i+=2)
{
res+=coef*i;
coef*=-1;
}
return res;
}
JAVA语言答第二题:
import java.util.Random;
public class CalculateFormulaValue {
public static void main(String[] args)
{
int[] array = new int[40];
array = generateArray(10);
for(int i=0; i<10; i++)
{
System.out.println(array[i]);
}
}
public static int[] generateArray(int n)
{
int[] res = new int[n];
Random randomGenerator = new Random();
for(int i=0; i
res[i]= randomGenerator.nextInt(50)+10;
}
return res;
}
}
不好意思了,原先看上面循环语句条件时我以为能解
结果看看JAVA语言。没办法,小弟我学C++的。。