（用java语言）编程把"1、2、3、4、5、6、7、8、9"分别放到下面的9个方框里使算式成立！

code=Java]public Demo1() {
int sum[] = new int[] { 1, 2, 3, 4, 5, 6, 7, 8, 9 }; for (int a = 0; a < 9; a++) {
for (int b = 0; b < 9; b++) {
for (int c = 0; c < 9; c++) {
for (int d = 0; d < 9; d++) {
for (int e = 0; e < 9; e++) {
for (int f = 0; f < 9; f++) {
for (int g = 0; g < 9; g++) {
for (int h = 0; h < 9; h++) {
for (int i = 0; i < 9; i++) {
if ((((sum[a] * 1000)
+ (sum[b] * 100)
+ (sum[c] * 10) + sum[d]) * sum[e]) == ((sum[f] * 1000)
+ (sum[g] * 100)
+ (sum[h] * 10) + sum[i])
&& bj(a, b, c, d, e, f, g,
h, i)) {
System.out.println(sum[a] + ""
+ sum[b] + sum[c]
+ sum[d] + "*" + sum[e]
+ "=" + sum[f] + sum[g]
+ sum[h] + sum[i]);
} }
}
}
}
} }
} } } } public boolean bj(int a, int b, int c, int d, int e, int f, int g, int h,
int i) {
boolean bean = true;
String ss[] = (a + "," + b + "," + c + "," + d + "," + e + "," + f
+ "," + g + "," + h + "," + i).split(",");
for (int j = 0; j < 9; j++) {
for (int k = 0; k < 9; k++) {
if (Integer.parseInt(ss[j]) == Integer.parseInt(ss[k])
&& j != k) {
bean = false;
}
}
}
return bean;
}[[/code]

package fillNum;import java.util.ArrayList;
import java.util.List;/**
* 产生排列数的方法
* @author t00132569
*
*/
public class Permutation {

public static List<List<Object>> generateCombination(List baseArray)
{

Object firstObj = baseArray.get(0);
if (baseArray.size() == 1)
{
List<List<Object>> result = new ArrayList<List<Object>>();
List<Object> inner = new ArrayList<Object>();
inner.add(firstObj);
result.add(inner);
return result;
}
baseArray.remove(0);
return combineWithArray(firstObj, baseArray);
}

private static List<List<Object>> combineWithArray(Object single, List<Object> otherArray)
{
List<List<Object>> otherCombineArray = generateCombination(otherArray);
List<List<Object>> finalResultList = new ArrayList<List<Object>>();
for (int i = 0; i < otherCombineArray.size(); i ++)
{
List<List<Object>> itemResult = new ArrayList<List<Object>>();
List<Object> combineItem = otherCombineArray.get(i);
/*减少计算, 倒着增加*/
for (int j = combineItem.size(); j >= 0; j--)
{
List<Object> sigleResult = new ArrayList<Object>();
sigleResult.addAll(combineItem);
sigleResult.add(j, single);
itemResult.add(sigleResult);
}
finalResultList.addAll(itemResult);
}
return finalResultList;
}
}

主算法:package fillNum;import java.util.ArrayList;
import java.util.List;public class Main {

private static boolean verify(List<Object> item)
{
StringBuffer first = new StringBuffer();
String second = null;
StringBuffer result = new StringBuffer();
for (int i = 0; i < item.size(); i++)
{
if (i <= 3 )
{
first.append(item.get(i));
}
if ( i == 4)
{
second = item.get(i).toString();
}

if (i > 4)
{
result.append(item.get(i));
}
}

int mul1 = Integer.valueOf(first.toString());
int mul2 = Integer.valueOf(second);
int finalRes = Integer.valueOf(result.toString());
if (finalRes == mul1 * mul2)
{
System.out.println(mul1 + " * " + mul2 + " = " + finalRes);
return true;
}
return false;
}

public static void main(String[] args)
{
List<String> stringList = new ArrayList<String>();
stringList.add("1");
stringList.add("2");
stringList.add("3");
stringList.add("4");
stringList.add("5");
stringList.add("6");
stringList.add("7");
stringList.add("8");
stringList.add("9");
List<List<Object>> list = Permutation.generateCombination(stringList);
for (int i = 0; i < list.size(); i++)
{
List<Object> item = list.get(i);
verify(item);
}

}}

    private static void count()
    {
     // the first one should be 1-4
     for (int a = 1; a < 5; a ++)
     {
     for (int b = 1; b <= 9; b ++)
     {
     if (a == b) continue; // the second one cann't equals to the first digit.

     for (int c = 1; c <=9 ; c ++)
     {
     if (c == a || c == b) continue; // the third one cann't euqals to the first, neither the second.

     for (int d = 2; d <=9 ; d++) // the fourth digit cann't be 1
     {
     if (d == a|| d == b|| d == c) continue; // ignore those digits that appeared

     // middle is between 2 - 8
     for (int middle = 2; middle <= 8; middle ++)
     {
     if (middle == a || middle == b || middle == c || middle == d) continue; // ignore those digits that appeared
     int first = a * 1000 + b * 100 + c * 10 + d;

     // the result should between 2000 - 9999
     int result = first * middle;
     if (result > 10000) break;

     // then count and find if that a good result.
     if (resultOK(first, middle, result))
     {
     System.out.println(first + " * " + middle + " = " + result);
     }
     }
     }
     }
     }
     }
    }

    private static boolean resultOK(int first, int middle, int result)
    {
     char [] firstC = String.valueOf(first).toCharArray();
     char [] resultC = String.valueOf(result).toCharArray();
     char [] finalC = new char[10];
     for (int i = 0; i < firstC.length; i ++)
     {
     finalC [firstC[i] - '0'] ++;
     }

     finalC[middle]++;

     for (int i = 0; i < resultC.length; i ++)
     {
     finalC [resultC[i] - '0'] ++;
     }

     if (finalC[0] > 0) return false;

     for (int i = 1; i < 10; i++)
     {
     if (finalC[i] != 1) return false;
     }

     return true;
    }
1738 * 4 = 6952
1963 * 4 = 7852

11楼的resultOK这个算法什么意思？

穷举法
import java.util.ArrayList;
import java.util.Date;
import java.util.List;public class Other {
public static void main(String args[]) {
long now = new Date().getTime();
String str = "123456789";
List list = new Other().getList(str);

// 遍历所有可能，将符合条件的数打印出来
for(int i = 0; i < list.size(); i++){
String num = list.get(i).toString();
int first = Integer.parseInt(num.substring(0, 4));
int second = Integer.parseInt(num.substring(4, 5));
int third = Integer.parseInt(num.substring(5, 9));

if(first * second == third){
System.out.println(first + " * " + second + " = " + third);
}
}
System.out.println("耗时：" + (new Date().getTime() - now) / 1000.0 + " 秒");
}

/**
* 得到字符串全排列
* @param str
* @return
*/
public List getList(String str){
List list = new ArrayList();

StringBuffer sb = new StringBuffer();
int leng = str.length();
int[] a = new int[leng];
boolean[] c = new boolean[leng];
for (int i = 0; i < leng; i++) {
a[i] = i + 1;
c[i] = false;
}
int ii = findActiveM(a, c); for (int i = 0; i < str.length(); i++){
sb.append("" + a[i]);
}
list.add(sb); if (ii >= 0) {
do {
int activeM = a[ii];
if (c[ii] == false) {
swapInt(a, ii, ii - 1);
swapBoolean(c, ii, ii - 1);
} else {
swapInt(a, ii, ii + 1);
swapBoolean(c, ii, ii + 1);
}
sb = new StringBuffer();
for (int i = 0; i < str.length(); i++) {
sb.append("" + a[i]);
if (a[i] > activeM)
c[i] = !c[i];
}
list.add(sb);
ii = findActiveM(a, c); } while (ii >= 0);
}

return list;
} private static void swapInt(int[] aa, int a, int b) {
int temp = aa[a];
aa[a] = aa[b];
aa[b] = temp;
} private static void swapBoolean(boolean[] aa, int a, int b) {
boolean temp = aa[a];
aa[a] = aa[b];
aa[b] = temp;
} private static int findActiveM(int[] a, boolean[] c) { int first1 = 0, first2 = 0;
int m = -1, n = -1;
for (int i = 1; i < a.length - 1; i++) {
if (c[i] == false) {
if ((a[i] > a[i - 1]) && a[i] > first1) {
first1 = a[i];
m = i;
}
} else {
if ((a[i] > a[i + 1]) && a[i] > first2) {
first2 = a[i];
n = i;
}
}
}
if ((c[0] == true) && (a[0] > a[1]) && (a[0] > first2)) {
first2 = a[0];
n = 0;
}
if ((c[a.length - 1] == false) && (a.length > 1)
&& (a[a.length - 1] > a[a.length - 2])
&& (a[a.length - 1] > first1)) {
first1 = a[a.length - 1];
m = a.length - 1;
}
if (first1 >= first2)
return m;
else
return n;
}
}

思路：穷举第二个乘数（k=?,2~9，1不可能）
然后从第一个乘数的个位开始赋值（=a），x=a*k,就可以得到结果的个位 x%10。同时结果的十位加上 x/10。
同理，再赋值乘数的十位(=b)，而这时的结果的十位应该 =b*k%10+(上面余留下来的 x/10)
赋值过程中，用 bArr[9] 对数字的使用情况进行判断。
    public static void g() {
        new Runnable() {            public void run() {
                for (k = 2; k <= 9; k++) {
                    bArr[k - 1] = true;
                    g(3);
                    bArr[k - 1] = false;
                }
            }            private void g(int c) {
                if (c < 0) {
                    for (int i = 0; i < aTop.length; i++) {
                        System.out.print(aTop[i] + " ");
                    }
                    System.out.println("\n" + k);
                    for (int i = 0; i < aBottom.length; i++) {
                        System.out.print(aBottom[i] + " ");
                    }
                    System.out.println("\n");
                    return;
                }
                for (int top = 1; top <= 9; top++) {
                    if (!bArr[top - 1]) {
                        bArr[top - 1] = true;
                        int x = top * k + aBottom[c];
                        int bottom = x % 10;
                        if (bottom > 0 && !bArr[bottom - 1] && (c > 0 || x < 10)) {
                            int y = aBottom[c];
                            aTop[c] = top;
                            aBottom[c] = bottom;
                            if (c > 0) {
                                aBottom[c - 1] = x / 10;
                            }
                            bArr[bottom - 1] = true;
                            g(c - 1);
                            bArr[bottom - 1] = false;
                            aBottom[c] = y;
                        }
                        bArr[top - 1] = false;
                    }
                }
            }
            private int[] aTop = new int[4];
            private int[] aBottom = new int[4];
            private boolean[] bArr = new boolean[9];
            private int k;
        }.run();
    }

遍历所有排列，再截字符串 public static void main(String[] args) {
for(int i=123456789; i<=987654321; i++) {
check(i);
}
} public static void check(int num) {
String str = String.valueOf(num);
if(str.contains("0"))
return;
for(int i=0; i<str.length(); i++) {
char c = str.charAt(i);
if(i != str.lastIndexOf(c))
return;
}

int num1 = Integer.parseInt(str.substring(0,4));
int num2 = str.charAt(4)-48;
int num3 = Integer.parseInt(str.substring(5));
if(num1 * num2 == num3) {
System.out.println(num1 + "×" + num2 + "=" + num3);
}
}

调试易

（用java语言）编程把"1、2、3、4、5、6、7、8、9"分别放到下面的9个方框里使算式成立！

解决方案 »