算法-大数相乘
2024-01-08 20:46:28
解决算法; * 1. 模拟小学乘法:最简单的乘法竖式手算的累加型; * 2. 分治乘法:最简单的是Karatsuba乘法,一般化以后有Toom-Cook乘法; * 3. 快速傅里叶变换FFT:(为了避免精度问题,可以改用快速数论变换FNTT),时间复杂度O(N lgN lglgN)。具体可参照Sch?nhage–Strassen algorithm; * 4. 中国剩余定理:把每个数分解到一些互素的模上,然后每个同余方程对应乘起来就行; * 5. Furer’s algorithm:在渐进意义上FNTT还快的算法。不过好像不太实用,本文就不作介绍了。大家可以参考维基百科Fürer’s algorithm
参考地址:https://blog.csdn.net/u010983881/article/details/77503519
1. 小学乘法代码如下:
/**
* 模拟小学乘法
*/
public static String primaryMul(String str1, String str2){
System.out.println("start primaryMul param=" + str1 + "," + str2 + ";time=" + System.currentTimeMillis());
int[] int1 = string2IntArray(str1);
int[] int2 = string2IntArray(str2);
List<Integer> result = new ArrayList<>();
for (int i = int1.length-1; i >=0 ; i--) {
List<Integer> tempList = new ArrayList<>();
int ten = 0;
for (int j = int2.length-1; j >=0 ; j--) {
int a = int1[i] * int2[j] + ten;
ten = a / 10;
tempList.add(a % 10);
}
if (ten > 0){
tempList.add(ten);
}
//两行相加
int carry = 0;
for (int j = 0; j < tempList.size(); j++) {
int r1 = 0;
int index = j + (int1.length-1-i);
if (index < 0 | result.size() > index){
r1 = result.get(index);
}
int a = tempList.get(j) + r1 + carry;
carry = a / 10;
if (index < 0 | result.size() > index){
result.set(index, a % 10);
}else{
result.add(a % 10);
}
}
if (carry > 0){
result.add(carry);
}
}
//方向旋转
int copy;
for (int i = 0; i < result.size(); i++) {
if (i < result.size() / 2){
copy = result.get(i);
int index = result.size() -1 -i;
result.set(i, result.get(index));
result.set(index, copy);
}
}
StringBuilder sb = new StringBuilder();
for (Integer integer : result) {
sb.append(integer);
}
System.out.println("end primaryMul result=" + sb + ";time=" + System.currentTimeMillis());
return sb.toString();
}
public static int[] string2IntArray(String str){
String[] strs = str.split("");
int[] ret = new int[strs.length];
for (int i = 0; i < strs.length; i++) {
ret[i] = Integer.parseInt(strs[i]);
}
return ret;
}
2. 小学算法-累加算法
?
/**
* 模拟小学乘法 - 累加算法
*/
private static String accumul(String str1, String str2){
System.out.println("start accumul param=" + str1 + "," + str2 + ";time=" + System.currentTimeMillis());
int[] int1 = string2IntArray(str1);
int[] int2 = string2IntArray(str2);
int[] result = new int[int1.length + int2.length];
for (int i = 0; i < int1.length; i++) {
for (int j = 0; j < int2.length; j++) {
result[i + j + 1] += int1[i] * int2[j];
}
}
for (int i = result.length - 1; i >= 0; i--) {
if (result[i] >= 10){
result[i-1] += result[i] / 10;
result[i] = result[i] % 10;
}
}
StringBuilder sb = new StringBuilder();
for (Integer integer : result) {
sb.append(integer);
}
System.out.println("end accumul result=" + sb + ";time=" + System.currentTimeMillis());
return sb.toString();
}
3. 分治算法-Karatsuba
/**
* Karatsuba乘法
*/
public static long karatsuba(long num1, long num2){
//递归终止条件
if(num1 < 10 || num2 < 10) return num1 * num2;
// 计算拆分长度
int size1 = String.valueOf(num1).length();
int size2 = String.valueOf(num2).length();
int halfN = Math.max(size1, size2) / 2;
/* 拆分为a, b, c, d */
long a = Long.valueOf(String.valueOf(num1).substring(0, size1 - halfN));
long b = Long.valueOf(String.valueOf(num1).substring(size1 - halfN));
long c = Long.valueOf(String.valueOf(num2).substring(0, size2 - halfN));
long d = Long.valueOf(String.valueOf(num2).substring(size2 - halfN));
// 计算z2, z0, z1, 此处的乘法使用递归
long z2 = karatsuba(a, c);
long z0 = karatsuba(b, d);
long z1 = karatsuba((a + b), (c + d)) - z0 - z2;
return (long)(z2 * Math.pow(10, (2*halfN)) + z1 * Math.pow(10, halfN) + z0);
}
文章来源:https://blog.csdn.net/aberwang9/article/details/135465148
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