英文:
How to add a condition so I don't go inside the loop
问题
我正在处理一段代码，在这段代码中，你需要从数组'b'中分别减去数组'a'的对应元素，直到数组'a'中的所有元素都相等为止。条件是对于减法操作，必须满足a[i] > b[i]。但并不是每次都能使所有元素相等，所以在这种情况下，我希望我的代码打印出'-1'，你可以无限制地进行多次减法操作。
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
short n = sc.nextShort();
short a[] = new short[n];
short b[] = new short[n];
for (short i = 0; i < n; i++) {// taking elements input
a[i] = sc.nextShort();
}
for (short i = 0; i < n; i++) {// taking elements input
b[i] = sc.nextShort();
}
short minimumOfArraya = 0;
for (short i = 0; i < n; i++) {// finding smallest element in array 'a'
for (short j = 0; j < n; j++) {
if (a[i] < a[j]) {
minimumOfArraya = a[i];
}
}
}
boolean allequal = false;
int counter = 0;
while (!allequal) {
for (short i = 0; i < n; i++) {// subtracting elements
if (a[i] == minimumOfArraya)
continue;
if (a[i] >= b[i]) {
a[i] -= b[i];
counter++;
}
}
for (short i = 0; i < n; i++) {
if (a[0] == a[i]) {
allequal = true;
} else {
allequal = false;
break;
}
}
}
for (int i = 0; i < n; i++) {// printing array 'a'
System.out.print(a[i] + " ");
}
System.out.println();
System.out.println(counter);
}
}
输入示例：
4
5 7 4 3
4 1 0 0
有效输入示例：
5
5 7 10 5 15
2 2 1 3 5
输出示例：
5 5 5 5 5
8
英文:
I am working on a code where you subtract respective elements of array 'a' from array'b' till you get all elements in array 'a' equal.<br />
Condition is that a[i]>b[i] for subtraction.<br />
But not every time getting all elements equal is possible, so in that case I would like my code to print '-1', how can I achieve it.<br /> I really tried hard to figure it out, don't give complex solutions as I am a beginner. you can subtract as many times you want.
Scanner sc = new Scanner(System.in);
short n = sc.nextShort();
short a[] = new short[n];
short b[] = new short[n];
for (short i = 0; i < n; i++) {// taking elements input
a[i] = sc.nextShort();
}
for (short i = 0; i < n; i++) {// taking elements input
b[i] = sc.nextShort();
}
short minimumOfArraya = 0;
for (short i = 0; i < n; i++) {// finding smallest element in array 'a'
for (short j = 0; j < n; j++) {
if (a[i] < a[j]) {
minimumOfArraya = a[i];
}
}
}
boolean allequal = false;
int counter = 0;
while (!allequal) {
for (short i = 0; i < n; i++) {// subtracting elements
if (a[i] == minimumOfArraya)
continue;
if (a[i] >= b[i]) {
a[i] -= b[i];
counter++;
}
}
for (short i = 0; i < n; i++) {
if (a[0] == a[i]) {
allequal = true;
} else {
allequal = false;
break;
}
}
}
for (int i = 0; i < n; i++) {// printing array 'a'
System.out.print(a[i] + " ");
}
System.out.println();
System.out.println(counter);
4
5 7 4 3//infinite loop
4 1 0 0
working input
5
5 7 10 5 15
2 2 1 3 5
output
5 5 5 5 5
8
答案1
得分: 0
快速找到最小值：
short minimumOfArraya = Short.MAX_VALUE;
for (short i = 0; i < n; i++) {// 在数组 'a' 中找到最小的元素
if (a[i] < minimumOfArraya) {
minimumOfArraya = a[i];
}
}
用于检查是否所有元素都相等的 for 循环一开始应该设置 allequal 为 true，
并且在找到 false 时中断。初始设置为 true 的部分遗漏了。
boolean allequal = false;
int counter = 0;
while (!allequal) {
for (short i = 0; i < n; i++) {// 减去元素
if (a[i] == minimumOfArraya)
continue;
if (a[i] >= b[i]) {
a[i] -= b[i];
counter++;
}
}
allequal = true;
for (short i = 0; i < n; i++) {
allequal = a[0] == a[i];
if (!allequal) {
break;
}
}
}
如果在 while 内部未增加 counter，代码可能会一直循环直到溢出。例如，如果最小值是 100，而 102 得到了 98。
英文:
The minimum can be found faster:
short minimumOfArraya = Short.MAX_VALUE;
for (short i = 0; i < n; i++) {// finding smallest element in array 'a'
if (a[i] < minimumOfArraya]) {
minimumOfArraya = a[i];
}
}
The for loop to check whether all are equal should initially have an allequal true,
and on finding a false, break. The initial setting to true was missing.
boolean allequal = false;
int counter = 0;
while (!allequal) {
for (short i = 0; i < n; i++) {// subtracting elements
if (a[i] == minimumOfArraya)
continue;
if (a[i] >= b[i]) {
a[i] -= b[i];
counter++;
}
}
allequal = true;
for (short i = 0; i < n; i++) {
allequal = a[0] == a[i];
if (!allequal) {
break;
}
}
}
If counter was not increased inside the while, the code may very well stay looping till overflow. If the minimum was 100 and 102 got 98 for instance.
答案2
得分: 0
如果一些迭代产生的数字小于最小值，这清楚地表明您无法使所有元素相等。在减去后检查。
while (!allequal) {
boolean impossible = false;
for (short i = 0; i < n; i++) {
if (a[i] < minimumOfArraya) {
impossible = true;
break;
}
if (a[i] == minimumOfArraya) continue;
if (a[i] >= b[i]) {
a[i] -= b[i];
counter++;
}
}
if (impossible) {
counter = -1;
break;
}
// 循环的其余部分
...
}
英文:
If some of iterations producing number which is less than minimum, it clearly shows that you cannot make all elements equal. Check that after subtracting
while (!allequal) {
boolean impossible = false;
for (short i = 0; i < n; i++) {
if (a[i] < mimimumofArraya) {
impossible = true;
break;
}
if (a[i] == minimumOfArraya) continue;
if (a[i] >= b[i]) {
a[i] -= b[i];
counter++;
}
}
if (impossible) {
counter = -1;
break;
}
// The rest of your loop
...
}
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