herrDeng網內搜尋
自訂搜尋
熱門文章
-
C++用 -1 mask法達到branchless解Leetcode1513 Number of Substrings With Only 1s|Py3 1-liner 其實很容易，想練功，試試一行解，或是把if分岔拿掉 [C++無分岔解請進]
-
請用C/C++ 程式計算 S=1*1+2*2+3*3+4*4+.....+20*20=?
-
計算你的BMI
-
計算你上學期加權平均
-
算上學期加權平均
-
GCD(123456789,345634567)=?
-
何謂演算法?並列出一無窮迴圈的範例!
-
55x+27y=gcd
-
問int怎麼存-689這個數字。 int用4個bytes。 答案請用16進位表達。
15 則留言:
public class hello {
public static void main(String[] args) {
double sum=0;
for(int n=1;n<1e3;n++)
sum=sum+1.0/(n*n);
System.out.print(sum);
public class ex1 {
public static void main(String[] args) {
double sum=0;
for(int n=1;n<1e3;n++)
sum=sum+1.0/(n*n);
System.out.print(sum);
public class ex1 {
public static void main(String[] args) {
double sum=0;
for(int n=1;n<1e3;n++)
sum=sum+1.0/(n*n);
System.out.print(sum);
public class EX
{
public static void main(String[] args)
{
double n,sum=0;
for(n=1;n<=1e6;n=n+1)
sum=sum+1.0/(n*n);
System.out.println(sum);
}
}
public class uilioloilil {
public static void main(String[] args) {
double sum=0;
for(int n=1;n<=1e3;n=n+1)
sum=1.0/(n*n)+sum;
System.out.println(sum);
}
package zzz;
public class ex1 {
public static void main(String[] args) {
int n;
double sum=0;
for(n=1;n<1e3;n=n+1)
sum=sum+1.0/(n*n);
System.out.println(sum);
}
}
public class aaa {
public static void main(String[] args) {
int n;
double sum=0;
for(n=1;n<=1e6;n=n+1)
sum=sum+1.0/((double)n*n);
System.out.println(sum);
}
}
public class ex1 {
public static void main(String[] args) {
int n;
double sum=0;
for(n=1;n<=1e3;n=n+1)
sum=sum+1.0/(n*n);
System.out.println(sum);
}
}
public class Hello {
public static void main(String[] args) {
double sum =0;
for(int n=1;n<=1e3;n++)
sum = sum + 1.0/(n*n);
System.out.print(sum);
}
}
public class ex2
{
public static void main(String[] args)
{
int n;
double sum=0;
for(n=1;n<1e3;n=n+1)
sum=sum+1.0/(n*n);
System.out.println(sum);
public class willia {
public static void main(String[] args) {
int n;
double sum=0;
for(n=1;n<=1e3;n++)
sum=sum+1.0/(n*n);
System.out.println(sum);
}
}
public class ex1 {
public static void main(String[] args) {
int n;
double sum=0;
for(n=1;n<=1e3;n=n+1)
sum=sum+1.0/(n*n);
System.out.println(sum);
}
}
public class ex1 {
public static void main(String[] args) {
int n;
double sum =0;
for(n=1;n<=1e3;n++)
sum=sum+1.0/(n*n);
System.out.println(sum);
}
public class ex2 {
public static void main(String[] args) {
int n;
double sun=0;
for(n=1;n<=1e3;n++)
sun=sun+1.0/(n*n);
System.out.println(sun);
}
}
public class ex1 {
public static void main(String[] args) {
int n;
double sun=0;
for(n=1;n<=1e3;n=n+1)
sun=sun+1.0/(n*n);
System.out.println(sun);
}
}
張貼留言