java代码实现线性代数 java中如何编写线程类

如何用JAVA程序解决三元一次方程组？

可以用线性代数java代码实现线性代数的知识解决。

把 X,Y,Z 和 后面的值 转化为 行列式，计算行列式的值就可以求解这道题目了。

叫 克莱姆法则，查查资料，就能解决了。

如果java代码实现线性代数你还需要答案的话，跟我联系。qqjava代码实现线性代数：2838844

相关问答

Q1: 谁帮我写一个多元一次方程自动求解的程序啊 用JAVA写

可以考虑线性代数多元一次方程解决方案

比如：

3x + 2y + 5z = 22

2x - 3y + 7z = 17

x + 6y - 3z = 4

则转化系数为行列式数组

int[][] param = {

{3, 2, 5, 22},

{2, -3, 7, 17},

{1, 6, -3, 4}

};

参考编码：

public static void main(String[] args) throws Exception {

int[][] param = {

{3, 2, 5, 22},

{2, -3, 7, 17},

{1, 6, -3, 4}

};

double[] result = new double[param.length];

double a = calc(param);

for (int i = 0; i param.length; i++) {

int[][] param2 = new int[param.length][param.length];

for (int m = 0; m param2.length; m++) {

for (int n = 0; n param2[m].length; n++) {

if (n == i) {

param2[m][n] = param[m][param[m].length - 1];

} else {

param2[m][n] = param[m][n];

}

}

}

result[i] = calc(param2) / a;

}

System.out.print("[");

for (int i = 0; i result.length; i++) {

if (i != 0) System.out.print(", ");

System.out.print(result[i]);

}

System.out.println("]");

}

private static int calc(int[][] param) {

int sum = 0;

int length = param.length;

for (int i = 0; i length; i++) {

int x = i;

int tmp = 1, tmp2 = 1;

for (int m = 0; m length; m++) {

tmp *= param[(x + m) % length][m];

tmp2 *= param[(x + m) % length][Math.abs(length - 1 - m) % length];

}

sum += tmp;

sum -= tmp2;

}

return sum;

}

Q2: 利用java实现2^1,2^2,2^3,2^4......,代码！

如果算的数小,可以放在int类型或者double类型的变量里.

太大的,就要放到BigInteger类型的变量里.

先看看放到int类型的变量代码

---------------------------------------------------------------

/*

* 数量比较小的情况下可以用这个来计算

* 太大了就存放不了了

*/

public class PowerNumber {

public static void main(String[] args) {

PowerNumber power = new PowerNumber();

// 准备求第几项的值

int n = 10;

// 这里先打印第4项

System.out.println("第" + n + "项结果是:" + power.powerNumberN(n));

// 这里打印前4项的和

System.out.println("前" + n + "项的和是:" + power.powerNumberSum(n));

}

// 计算第n项的值

int powerNumberN(int n) {

// 用来存放第n个数

int sum = 1;

for (int i = 1; i = n; i++) {

sum = 2 * sum;

}

return sum;

}

// 计算前n项的和

int powerNumberSum(int n) {

// 存放前n项的和

int sumNum = 0;

int frontN = n;

for (int i = 1; i = n; i++) {

// 存放第n项的值

int sum = 1;

for (int j = 1; j = frontN; j++) {

// 第n项的值

sum = 2 * sum;

}

//求前一项

frontN--;

// 求出的值加到总和里

sumNum = sumNum + sum;

}

return sumNum;

}

}

------------------------------------------------------

前4项的结果:

第4项结果是:16

前4项的和是:30

--------------------

前20项的结果:

第20项结果是:1048576

前20项的和是:2097150

----------------------

前40项的结果......

第40项结果是:0

前40项的和是:-2

================================================

再看看放到BigInteger类型的变量里的程序

---------------------------------

import java.math.BigInteger;

public class Power {

public static void main(String[] args) {

Power power = new Power();

// 准备求第几项的值

int n = 4;

// 这里先打印第4项

System.out.println("第" + n + "项结果是:" + power.powerNum(n));

// 这里打印前4项的和

System.out.println("前" + n + "项的和是:" + power.sumPowerNum(n));

}

// 计算第n项的值

BigInteger powerNum(int n) {

BigInteger num = BigInteger.ONE;

for (int i = 1; i = n; i++) {

// 为了防止结果过大,将结果放在BigInteger中

// 每次对结果乘以2

num = num.multiply(new BigInteger(new Integer(2).toString()));

}

// 打印2^n结果

return num;

}

// 计算前n项的值

BigInteger sumPowerNum(int n) {

// 存放前n项的值

BigInteger sumNum = BigInteger.ZERO;

int frontN = n;

for (int i = 1; i = n; i++) {

// 存放第n项的值

BigInteger num = BigInteger.ONE;

for (int j = 1; j = frontN; j++) {

// 为了防止结果过大,将结果放在BigInteger中

// 每次对结果乘以2

num = num.multiply(new BigInteger(new Integer(2).toString()));

}

// 每次循环让最大值减掉1,以计算前面的值

frontN--;

// 计算出第n项的值,将其放入总和sumNum中

sumNum = sumNum.add(num);

}

return sumNum;

}

}

---------------------------------------------------

前4项的结果:

第4项结果是:16

前4项的和是:30

--------------------------------------------------

前40项的结果:

第40项结果是:1099511627776

前40项的和是:2199023255550

---------------------------------------------------------------------------------------------

前400项的结果:

第400项结果是:2582249878086908589655919172003011874329705792829223512830659356540647622016841194629645353280137831435903171972747493376

前400项的和是:5164499756173817179311838344006023748659411585658447025661318713081295244033682389259290706560275662871806343945494986750

-------------------------------------------------------------------------------------

前4000项的结果:

第4000项结果是:13182040934309431001038897942365913631840191610932727690928034502417569281128344551079752123172122033140940756480716823038446817694240581281731062452512184038544674444386888956328970642771993930036586552924249514488832183389415832375620009284922608946111038578754077913265440918583125586050431647284603636490823850007826811672468900210689104488089485347192152708820119765006125944858397761874669301278745233504796586994514054435217053803732703240283400815926169348364799472716094576894007243168662568886603065832486830606125017643356469732407252874567217733694824236675323341755681839221954693820456072020253884371226826844858636194212875139566587445390068014747975813971748114770439248826688667129237954128555841874460665729630492658600179338272579110020881228767361200603478973120168893997574353727653998969223092798255701666067972698906236921628764772837915526086464389161570534616956703744840502975279094087587298968423516531626090898389351449020056851221079048966718878943309232071978575639877208621237040940126912767610658141079378758043403611425454744180577150855204937163460902512732551260539639221457005977247266676344018155647509515396711351487546062479444592779055555421362722504575706910949376

前4000项的和是:26364081868618862002077795884731827263680383221865455381856069004835138562256689102159504246344244066281881512961433646076893635388481162563462124905024368077089348888773777912657941285543987860073173105848499028977664366778831664751240018569845217892222077157508155826530881837166251172100863294569207272981647700015653623344937800421378208976178970694384305417640239530012251889716795523749338602557490467009593173989028108870434107607465406480566801631852338696729598945432189153788014486337325137773206131664973661212250035286712939464814505749134435467389648473350646683511363678443909387640912144040507768742453653689717272388425750279133174890780136029495951627943496229540878497653377334258475908257111683748921331459260985317200358676545158220041762457534722401206957946240337787995148707455307997938446185596511403332135945397812473843257529545675831052172928778323141069233913407489681005950558188175174597936847033063252181796778702898040113702442158097933437757886618464143957151279754417242474081880253825535221316282158757516086807222850909488361154301710409874326921805025465102521079278442914011954494533352688036311295019030793422702975092124958889185558111110842725445009151413821898750

Q3: 要编写一个两个矩阵相乘的JAVA方法，本人不懂，求高手帮忙！！万分感激！！

你好，按照你的要求代码如下，给出了注释和运行结果，可以直接运行：

public class test2 {

public static int[][] multiplyMatrix(int[][] a, int[][] b) {

// 判断是否合法

if (a == null || a == null || a.length == 0 || b.length == 0

|| a[0].length != b.length) {

return null;

}

// 计算相乘

int[][] c = new int[a.length][b[0].length];

for (int i = 0; i a.length; i++) {

for (int j = 0; j b[0].length; j++) {

for (int k = 0; k a[0].length; k++) {

c[i][j] += a[i][k] * b[k][j];

}

}

}

return c;

}

public static void main(String[] args) {

int[][] a = new int[][] { { 1, 2, 3 }, { 1, 2, 3 } };

int[][] b = new int[][] { { 1, 2 }, { 1, 2 }, { 1, 2 } };

int[][] c = multiplyMatrix(a, b);

printMatrix(a);

printMatrix(b);

printMatrix(c);

}

// 打印矩阵

public static void printMatrix(int[][] c) {

if (c != null) {

for (int i = 0; i c.length; i++) {

for (int j = 0; j c[0].length; j++) {

System.out.print(c[i][j] + " ");

}

System.out.println();

}

} else {

System.out.println("无效");

}

System.out.println();

}

}

运行结果：

1 2 3

1 2 3

1 2

1 2

1 2

6 12

6 12

Q4: java代码如何实现a^-1(mod b)？

（1）用循环一个个试

（2）要快一点java代码实现线性代数，试试用扩展欧几里得算法（就是利用辗转相除找最大公因数java代码实现线性代数，算法过程中记录一些东西java代码实现线性代数，就知道是否有逆，有逆直接可求出来），

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