如果算的数小,可以放在int类型或者double类型的变量里.
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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
这个问题，我好像回答过！
/**
* 最小二乘法计算类
*
* @author Administrator
*
*/
public class LeastSquareMethod {
private double[] x;
private double[] y;
private double[] weight;
private int m;
private double[] coefficient;
public LeastSquareMethod(double[] x, double[] y, int m) {
if (x == null || y == null || x.length 2 || x.length != y.length
|| m 2)
throw new IllegalArgumentException("无效的参数");
this.x = x;
this.y = y;
this.m = m;
weight = new double[x.length];
for (int i = 0; i x.length; i++) {
weight[i] = 1;
}
}
public LeastSquareMethod(double[] x, double[] y, double[] weight, int m) {
if (x == null || y == null || weight == null || x.length 2
|| x.length != y.length || x.length != weight.length || m 2)
throw new IllegalArgumentException("无效的参数");
this.x = x;
this.y = y;
this.m = m;
this.weight = weight;
}
public double[] getCoefficient() {
if (coefficient == null)
compute();
return coefficient;
}
public double fit(double v) {
if (coefficient == null)
compute();
if (coefficient == null)
return 0;
double sum = 0;
for (int i = 0; i coefficient.length; i++) {
sum += Math.pow(v, i) * coefficient[i];
}
return sum;
}
private void compute() {
if (x == null || y == null || x.length = 1 || x.length != y.length
|| x.length m || m 2)
return;
double[] s = new double[(m - 1) * 2 + 1];
for (int i = 0; i s.length; i++) {
for (int j = 0; j x.length; j++)
s[i] += Math.pow(x[j], i) * weight[j];
}
double[] f = new double[m];
for (int i = 0; i f.length; i++) {
for (int j = 0; j x.length; j++)
f[i] += Math.pow(x[j], i) * y[j] * weight[j];
}
double[][] a = new double[m][m];
for (int i = 0; i m; i++) {
for (int j = 0; j m; j++) {
a[i][j] = s[i + j];
}
}
coefficient = Algorithm.multiLinearEquationGroup(a, f);
}
/**
* @param args
*/
public static void main(String[] args) {
LeastSquareMethod l = new LeastSquareMethod(
new double[] { 2000, 2001, 2002, 2003, 2004, 2005, 2006, 2007, 2008 },
new double[] { 37.84, 44.55, 45.74, 63.8, 76.67, 105.59, 178.48, 355.27, 409.92 },
new double[] { 11, 12, 13, 14, 15, 16, 17, 18, 19 },
2);
double[] x = l.getCoefficient();
for (double xx : x) {
System.out.println(xx);
}
System.out.println(l.fit(2009));
}
}
最小二乘发拟合的是直线吧，不是曲线，非线性曲线拟合你的有模型，知道曲线的方程式。或者就是差值了，三次样条或者B样条。
因为知道多组数据，就是按照矩阵的方式来求.建立矩阵乘法之后，通过在两边同时乘以原矩阵的逆矩阵，然后就能得到三元方程组的解了.如果不会的话，可以上网查一下java实现矩阵乘法以及java实现逆矩阵..然后你的这个功能就完成了.
肯定的啊.第二个程序循环
for (int j=1;j==i;j++){
System.out.print(i+"*"+j+"="+(i*j)+"\t");
}
i=1时,j=1,好吧,出来了1*1=1
j=2时,i==j不成立了,所以j不++了.所以j永远是2了.永远不等于,所以不会打印了.
i=2,3,4,5,6,7,8,9时
j开始等于1,结果j永远不会等于i,所以j永远是1了,后面的也就不会执行,不会打印了
我是高级Java从业者，数学功底一般般，不给你翻译代码了。这类问题一般用matlab的多，用java处理计算问题的比较少。你可以参考国外大学的一些在线研究成果。如MIT的很多项目都直接放在网上。
他山之石，可以攻玉。这里给你找了一个模拟拟合的网站：
