Java 二次方程系数计算
关于二次方程（了解更多），我将方程的Java 二次方程系数计算,java,quadratic,Java,Quadratic,关于二次方程（了解更多），我将方程的a、b和c作为输入 示例公式如下：21x^2-8x-4 这里，a=21，b=-8，c=-4。因此，在求解（无公式）时， =>21x^2-14x+6x-4=0 我需要两个中间数字，也就是说，在这个例子中是14和6（读取因子）。我认为我所做的一切都是正确的，但输入似乎是无限的，而且一点也没有停止。你能纠正这个错误吗？我也很想知道为什么会这样 import java.util.Scanner; public class QuadFact { static S
abcimport java.util.Scanner;
public class QuadFact {
static Scanner sc = new Scanner(System.in);
static int a,b,c;
static int P, diff, p;
static int i;
static boolean found = false;
void accept(){
System.out.println("Enter the a, b, c");
a = sc.nextInt(); b = sc.nextInt(); c = sc.nextInt();
}
void compute(){
P = a * c;
diff = 0;
while(!found){
for (i = b + 1;;i++){
diff = i - b;
p = i * diff;
if (p==P) {
found = true;
break;
}
}
}
}
void display(){
System.out.print("These are the raw numbers, should be correct.
Still,\n it is advisable you verify it.");
System.out.println("One factor: " + i);
System.out.println("Other factor: " + diff);
}
public static void main(String[] args){
QuadFact a = new QuadFact();
a.accept();
a.compute();
a.display();
}
}
void compute(){
P = a * c;
while(!found){
for( i = 1; ; i++ ){
diff = b - i;
if (i * diff == P) {
found = true;
break;
}
diff = b + i;
if (-i * diff == P) {
found = true;
break;
}
}
}
}
void compute(){
P = a * c;
while(!found){
for( i = 1; ; i++ ){
diff = b - i;
if (i * diff == P) {
found = true;
break;
}
diff = b + i;
if (-i * diff == P) {
found = true;
break;
}
}
}
}
void compute(){
P = a * c;
while(!found){
for( i = 1; ; i++ ){
diff = b - i;
if (i * diff == P) {
found = true;
break;
}
diff = b + i;
if (-i * diff == P) {
found = true;
break;
}
}
}
}
void compute(){
P = a * c;
while(!found){
for( i = 1; ; i++ ){
diff = b - i;
if (i * diff == P) {
found = true;
break;
}
diff = b + i;
if (-i * diff == P) {
found = true;
break;
}
}
}
}
好的，我为此写了一个代码
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
// TODO Auto-generated method stub
//Declare and get the variables
int a, b,c;
Scanner s = new Scanner(System.in);
System.out.println("Enter A");
a = s.nextInt();
System.out.println("Enter B");
b = s.nextInt();
System.out.println("Enter c");
c = s.nextInt();
//A should be 1 if not divide a, b and c by a
if (a>1) {
b= b/a;
c=c/a;
a= a/a;
}
//Just printing what the values of ABC IS AGAIN
System.out.println("A = "+a+" B = "+b+" C = "+c);
//Just printing what the values of ABC IS AGAIN but in reverse, that is if c was 5 it becomes -5
//another way to reverse positive to negative and vice versa
System.out.println("A = "+(0-a)+" B = "+(0-b)+" C = "+(0-c));
//Set i as c and start the loop from highest to lowest.
for (int i = Math.abs(c); i>0 ;i-- ) {
//if a multiple is found it proceeds and checks for the
//multiple combination that when multiplied you get C and the addition or subtraction gives you B
if (c%i==0) {
int fac1 = c/i;
//Displays the multiples found
System.out.println(i+" x "+fac1+" = "+c);
//There are 4 possible outcomes or cases
//case 1 multiple 1 - multiple 2 = b
//case 2 multiple 2 - multiple 1 = b
//case 3 multiple 1 + multiple 2 = b
//case 4 -multiple 1 + -multiple 2 = b
if (i-fac1 == b) {
//System.out.println("case 1: " + i+"-"+fac1+"="+b);
answer(i,fac1);
break;
}
else if(fac1 - i == b){
//System.out.println("case2: " + fac1+"-"+i+"="+b);
answer(fac1,i);
break;
}
else if (i +fac1 == b ) {
//System.out.println("case3: " + i+"+"+fac1+"="+b);
answer(i,fac1);
break;
}
else if((0-Math.abs(i)) + (0- Math.abs(fac1))==b) {
//System.out.println("case4: "+"-"+i+ " + "+ "-"+ fac1 +"="+b);
answer((0-Math.abs(i)),(0- Math.abs(fac1)));
break;
}
else {
System.out.println("Probably not a factorizable Equation");
}
}
}
}
//Use this method to show the final answer
private static void answer(int f1, int f2) {
System.out.println("x = "+(0-f1) +" or x = "+(0-f2));
}
}
好的，我为此写了一个代码
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
// TODO Auto-generated method stub
//Declare and get the variables
int a, b,c;
Scanner s = new Scanner(System.in);
System.out.println("Enter A");
a = s.nextInt();
System.out.println("Enter B");
b = s.nextInt();
System.out.println("Enter c");
c = s.nextInt();
//A should be 1 if not divide a, b and c by a
if (a>1) {
b= b/a;
c=c/a;
a= a/a;
}
//Just printing what the values of ABC IS AGAIN
System.out.println("A = "+a+" B = "+b+" C = "+c);
//Just printing what the values of ABC IS AGAIN but in reverse, that is if c was 5 it becomes -5
//another way to reverse positive to negative and vice versa
System.out.println("A = "+(0-a)+" B = "+(0-b)+" C = "+(0-c));
//Set i as c and start the loop from highest to lowest.
for (int i = Math.abs(c); i>0 ;i-- ) {
//if a multiple is found it proceeds and checks for the
//multiple combination that when multiplied you get C and the addition or subtraction gives you B
if (c%i==0) {
int fac1 = c/i;
//Displays the multiples found
System.out.println(i+" x "+fac1+" = "+c);
//There are 4 possible outcomes or cases
//case 1 multiple 1 - multiple 2 = b
//case 2 multiple 2 - multiple 1 = b
//case 3 multiple 1 + multiple 2 = b
//case 4 -multiple 1 + -multiple 2 = b
if (i-fac1 == b) {
//System.out.println("case 1: " + i+"-"+fac1+"="+b);
answer(i,fac1);
break;
}
else if(fac1 - i == b){
//System.out.println("case2: " + fac1+"-"+i+"="+b);
answer(fac1,i);
break;
}
else if (i +fac1 == b ) {
//System.out.println("case3: " + i+"+"+fac1+"="+b);
answer(i,fac1);
break;
}
else if((0-Math.abs(i)) + (0- Math.abs(fac1))==b) {
//System.out.println("case4: "+"-"+i+ " + "+ "-"+ fac1 +"="+b);
answer((0-Math.abs(i)),(0- Math.abs(fac1)));
break;
}
else {
System.out.println("Probably not a factorizable Equation");
}
}
}
}
//Use this method to show the final answer
private static void answer(int f1, int f2) {
System.out.println("x = "+(0-f1) +" or x = "+(0-f2));
}
}
好的，我为此写了一个代码
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
// TODO Auto-generated method stub
//Declare and get the variables
int a, b,c;
Scanner s = new Scanner(System.in);
System.out.println("Enter A");
a = s.nextInt();
System.out.println("Enter B");
b = s.nextInt();
System.out.println("Enter c");
c = s.nextInt();
//A should be 1 if not divide a, b and c by a
if (a>1) {
b= b/a;
c=c/a;
a= a/a;
}
//Just printing what the values of ABC IS AGAIN
System.out.println("A = "+a+" B = "+b+" C = "+c);
//Just printing what the values of ABC IS AGAIN but in reverse, that is if c was 5 it becomes -5
//another way to reverse positive to negative and vice versa
System.out.println("A = "+(0-a)+" B = "+(0-b)+" C = "+(0-c));
//Set i as c and start the loop from highest to lowest.
for (int i = Math.abs(c); i>0 ;i-- ) {
//if a multiple is found it proceeds and checks for the
//multiple combination that when multiplied you get C and the addition or subtraction gives you B
if (c%i==0) {
int fac1 = c/i;
//Displays the multiples found
System.out.println(i+" x "+fac1+" = "+c);
//There are 4 possible outcomes or cases
//case 1 multiple 1 - multiple 2 = b
//case 2 multiple 2 - multiple 1 = b
//case 3 multiple 1 + multiple 2 = b
//case 4 -multiple 1 + -multiple 2 = b
if (i-fac1 == b) {
//System.out.println("case 1: " + i+"-"+fac1+"="+b);
answer(i,fac1);
break;
}
else if(fac1 - i == b){
//System.out.println("case2: " + fac1+"-"+i+"="+b);
answer(fac1,i);
break;
}
else if (i +fac1 == b ) {
//System.out.println("case3: " + i+"+"+fac1+"="+b);
answer(i,fac1);
break;
}
else if((0-Math.abs(i)) + (0- Math.abs(fac1))==b) {
//System.out.println("case4: "+"-"+i+ " + "+ "-"+ fac1 +"="+b);
answer((0-Math.abs(i)),(0- Math.abs(fac1)));
break;
}
else {
System.out.println("Probably not a factorizable Equation");
}
}
}
}
//Use this method to show the final answer
private static void answer(int f1, int f2) {
System.out.println("x = "+(0-f1) +" or x = "+(0-f2));
}
}
好的，我为此写了一个代码
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
// TODO Auto-generated method stub
//Declare and get the variables
int a, b,c;
Scanner s = new Scanner(System.in);
System.out.println("Enter A");
a = s.nextInt();
System.out.println("Enter B");
b = s.nextInt();
System.out.println("Enter c");
c = s.nextInt();
//A should be 1 if not divide a, b and c by a
if (a>1) {
b= b/a;
c=c/a;
a= a/a;
}
//Just printing what the values of ABC IS AGAIN
System.out.println("A = "+a+" B = "+b+" C = "+c);
//Just printing what the values of ABC IS AGAIN but in reverse, that is if c was 5 it becomes -5
//another way to reverse positive to negative and vice versa
System.out.println("A = "+(0-a)+" B = "+(0-b)+" C = "+(0-c));
//Set i as c and start the loop from highest to lowest.
for (int i = Math.abs(c); i>0 ;i-- ) {
//if a multiple is found it proceeds and checks for the
//multiple combination that when multiplied you get C and the addition or subtraction gives you B
if (c%i==0) {
int fac1 = c/i;
//Displays the multiples found
System.out.println(i+" x "+fac1+" = "+c);
//There are 4 possible outcomes or cases
//case 1 multiple 1 - multiple 2 = b
//case 2 multiple 2 - multiple 1 = b
//case 3 multiple 1 + multiple 2 = b
//case 4 -multiple 1 + -multiple 2 = b
if (i-fac1 == b) {
//System.out.println("case 1: " + i+"-"+fac1+"="+b);
answer(i,fac1);
break;
}
else if(fac1 - i == b){
//System.out.println("case2: " + fac1+"-"+i+"="+b);
answer(fac1,i);
break;
}
else if (i +fac1 == b ) {
//System.out.println("case3: " + i+"+"+fac1+"="+b);
answer(i,fac1);
break;
}
else if((0-Math.abs(i)) + (0- Math.abs(fac1))==b) {
//System.out.println("case4: "+"-"+i+ " + "+ "-"+ fac1 +"="+b);
answer((0-Math.abs(i)),(0- Math.abs(fac1)));
break;
}
else {
System.out.println("Probably not a factorizable Equation");
}
}
}
}
//Use this method to show the final answer
private static void answer(int f1, int f2) {
System.out.println("x = "+(0-f1) +" or x = "+(0-f2));
}
}
再次尝试在此处发布您的代码。我们可以帮你格式化。链接可能会失效，我们中的一些人（比如我自己）甚至无法访问您的链接。好吧，当您尝试使用调试器调试此链接时会发生什么情况，或者当您使用println语句查看您的代码在关键点执行什么操作时会发生什么情况？您已经弄糊涂了。在你的例子中，你想要的数字是-14和6，而不是14和6。你应该寻找两个乘积为P和和为b的数字，而你似乎在寻找乘积为P和差为b的数字。输入是无限量，再次在这里发布你的代码。我们可以帮你格式化。链接可能会失效，我们中的一些人（比如我自己）甚至无法访问您的链接。好吧，当您尝试使用调试器调试此链接时会发生什么情况，或者当您使用println语句查看您的代码在关键点执行什么操作时会发生什么情况？您已经弄糊涂了。在你的例子中，你想要的数字是-14和6，而不是14和6。你应该寻找两个乘积为P和和为b的数字，而你似乎在寻找乘积为P和差为b的数字。输入是无限量，再次在这里发布你的代码。我们可以帮你格式化。链接可能会失效，我们中的一些人（比如我自己）甚至无法访问您的链接。好吧，当您尝试使用调试器调试此链接时会发生什么情况，或者当您使用println语句查看您的代码在关键点执行什么操作时会发生什么情况？您已经弄糊涂了。在你的例子中，你想要的数字是-14和6，而不是14和6。你应该寻找两个乘积为P和和为b的数字，而你似乎在寻找乘积为P和差为b的数字。输入是无限量，再次在这里发布你的代码。我们可以帮你格式化。链接可能会失效，我们中的一些人（比如我自己）甚至无法访问您的链接。好吧，当您尝试使用调试器调试此链接时会发生什么情况，或者当您使用println语句查看您的代码在关键点执行什么操作时会发生什么情况？您已经弄糊涂了。在你的例子中，你想要的数字是-14和6，而不是14和6。你应该寻找两个乘积为P和和为b的数字，而你似乎在寻找乘积为P和差为b的数字。输入是无穷大的。这通常在a=1时有效，并且可以被c或b平均除。在类似三项式的情况下，它可能不起作用。注意，这主要是在a=1时起作用，并且可以被c或b平均分割。在类似三项式的情况下，它可能不起作用。注意，这主要是在a=1时起作用，并且可以被c或b平均分割。在类似三项式的情况下，它可能不起作用。注意，这主要是在a=1时起作用，并且可以被c或b平均分割。在类似三项式的情况下，它可能不起作用