分治法——大数相乘 大数相乘（分治法）

大数相乘（分治法）

前言：

A,B两个大数，都为n位，要计算A*B，需要将A和B划分成两等份，如下图所示

普通的做法是A*B=a1*b1*10^n+(a1*b0+b1*a0）*10^(2/n)+a0*b0

举个例子：

1234*9876=（12*98）*10000+（12*76+98*34）*100+34*76

对于这个算法的时间复杂度，我们需要做4次n/2级别的乘法和3加法。即T(n)=4*T(n/2)+O(n),时间复杂度是O(n²）.

分治法的算法是A*B=a1*b1*10^n+[(a1+a0)*(b0+b1)-a1*a0-b1*b0]*10^n/2+a0*b0

对于这个算法的时间复杂度，我们需要做3次n/2级别的乘法。即T(n)=3*T(n/2)+O(n),时间复杂度是T(n) = O(n^log2(3) ) = O(n^1.59 ).

下面上代码：

     
     
#include<iostream>
#include<string>
#include<algorithm>
#include<sstream>
using namespace std;
//string转换为int
int strint(string s) {
int num;
stringstream ss(s);
ss >> num;
return num;
}
//int转换为string
string intstr(int num) {
string s;
stringstream ss;
ss << num;
ss >> s;
return s;
}
//前置0 ，因为大数的长度必须要为2^n的倍数，才能划分
void removePrezero(string &s) {
if (s.length() == 1) return;
int k = 0;
for ( int i = 0; i < s.length(); i++) {
if (s[i] == '0')k++;
else break;
}
if (k == s.length())s = "0";
else s = s.substr(k);
}
//大数相加
string add(string s1, string s2) {
string s = "";
removePrezero(s1); //去掉前置0
removePrezero(s2);
reverse(s1.begin(), s1.end()); //先把字符串颠倒，方便相加
reverse(s2.begin(), s2.end());
int c = 0;
for ( int i = 0;c|| i < max(s1.length(), s2.length()); i++) {
int t = c;
if (i < s1.length())t += s1[i] - '0';
if (i < s2.length())t += s2[i] - '0';
int d = t % 10;
s.insert( 0, intstr(d));
c = t / 10;
}
return s;
}
//大数相减
string subtra(string &s1, string &s2) {
string s = "";
string flag;
removePrezero(s1); //去掉前置0
removePrezero(s2);
int len1 = s1.length();
int len2 = s2.length();
int len = len1>len2 ? len1 : len2;
if (len1 < len2)flag = "-";
else if (len1 > len2)flag = "+";
else {
int i;
for (i = 0; i < len1; i++) {
if (s1.at(i) > s2.at(i)) {
flag = "+"; break;
}
else if (s1.at(i) < s2.at(i)) {
flag = "-"; break;
}
}
if (i == len1)s == "0";
}
int* num = ( int*) malloc( sizeof( int)*len);
reverse(s1.begin(), s1.end()); //方便相减
reverse(s2.begin(), s2.end());
int c = 0;
for ( int j = 0; j < len; j++) {
int n1 = j < len1 ? s1[j] - '0' : 0;
int n2 = j < len2 ? s2[j] - '0' : 0;
if (flag == "+")num[c++] = n1 - n2;
else num[c++] = n2 - n1;
}
for ( int j = 0; j < c; j++) {
if (num[j] < 0) {
num[j] += 10; num[j + 1] -= 1;
}
}
c--;
while (num[c] == 0)c--;
for ( int j = 0; j <=c; j++){
s.insert( 0, intstr(num[j]));
}
if (flag == "-") return flag + s;
else return s;
}
//增加前置0
void addPrezero(string &s, int L) {
for ( int i = 0; i < L; i++)
s = s.insert( 0, "0");
}
//增加后置0，大数*10^n 相当于在数的后面加n个0
string addLastzero(string s, int L) {
string s1 = s;
for ( int i = 0; i < L; i++)
s1 += "0";
return s1;
}
//大数相乘
string multi(string &s1, string &s2) {
bool flag1 = false, flag2 = false;
string sign; //作用是判断结果是正数还是负数
if (s1.at( 0) == '-') {
flag1 = true; s1 = s1.substr( 1);
}
if (s1.at( 0) == '-') {
flag2 = true; s2 = s2.substr( 1);
}
if (flag1&&flag2)sign = "+";
else if (flag1 || flag2) sign = "-";
else sign = "+";
int L = 4; //将数前置若干个0，使其长度为2^n的倍数
if (s1.length() > 2 || s2.length() > 2) {
if (s1.length() >= s2.length()) {
while (L < s1.length())L *= 2;
if (L != s1.length())
addPrezero(s1, L - s1.length());
addPrezero(s2, L - s2.length());
} else {
while (L < s2.length())L*= 2;
if (L != s2.length())
addPrezero(s2, L - s2.length());
addPrezero(s1, L - s1.length());
}
}
if (s1.length() == 1)addPrezero(s1, 1);
if (s2.length() == 1)addPrezero(s2, 1);

int n = s1.length();
string result, a0, a1, b0, b1;
if (n > 1) {
a1 = s1.substr( 0, n / 2);
a0 = s1.substr(n / 2, n);
b1 = s2.substr( 0, n / 2);
b0 = s2.substr(n / 2, n);
}
if (n == 2) {
int x1 = strint(a1);
int x2 = strint(a0);
int y1 = strint(b1);
int y2 = strint(b0);
int num = (x1 * 10 + x2)*(y1 * 10 + y2);
result = intstr(num);
} else {
string c2 = multi(a1, b1);
string c0 = multi(a0, b0);
string temp1 = add(a0, a1);
string temp2 = add(b1, b0);
string temp3 = add(c2, c0);
string temp_c1 = multi(temp1, temp2);
string c1 = subtra(temp_c1, temp3);
string s1 = addLastzero(c1, n / 2);
string s2 =addLastzero(c2, n);
result = add(add(s1, s2), c0);
}
if (sign == "-")result.insert( 0, sign);
return result;
}
int main() {
//text
string s1 = "-1234999999999999999999999", s2 = "12348976543";
cout << multi(s1, s2);
}
     
     

总结：算法本身不难，主要是程序有太多需要注意的细节。对代码有疑问或者有些部分不理解的朋友，欢迎留言。

对了，程序的主要思想参考于点击打开链接，强烈建议大家可以去看一下。

前言：

A,B两个大数，都为n位，要计算A*B，需要将A和B划分成两等份，如下图所示

普通的做法是A*B=a1*b1*10^n+(a1*b0+b1*a0）*10^(2/n)+a0*b0

举个例子：

1234*9876=（12*98）*10000+（12*76+98*34）*100+34*76

对于这个算法的时间复杂度，我们需要做4次n/2级别的乘法和3加法。即T(n)=4*T(n/2)+O(n),时间复杂度是O(n²）.

分治法的算法是A*B=a1*b1*10^n+[(a1+a0)*(b0+b1)-a1*a0-b1*b0]*10^n/2+a0*b0

对于这个算法的时间复杂度，我们需要做3次n/2级别的乘法。即T(n)=3*T(n/2)+O(n),时间复杂度是T(n) = O(n^log2(3) ) = O(n^1.59 ).

下面上代码：

  
  
#include<iostream>
#include<string>
#include<algorithm>
#include<sstream>
using namespace std;
//string转换为int
int strint(string s) {
int num;
stringstream ss(s);
ss >> num;
return num;
}
//int转换为string
string intstr(int num) {
string s;
stringstream ss;
ss << num;
ss >> s;
return s;
}
//前置0 ，因为大数的长度必须要为2^n的倍数，才能划分
void removePrezero(string &s) {
if (s.length() == 1) return;
int k = 0;
for ( int i = 0; i < s.length(); i++) {
if (s[i] == '0')k++;
else break;
}
if (k == s.length())s = "0";
else s = s.substr(k);
}
//大数相加
string add(string s1, string s2) {
string s = "";
removePrezero(s1); //去掉前置0
removePrezero(s2);
reverse(s1.begin(), s1.end()); //先把字符串颠倒，方便相加
reverse(s2.begin(), s2.end());
int c = 0;
for ( int i = 0;c|| i < max(s1.length(), s2.length()); i++) {
int t = c;
if (i < s1.length())t += s1[i] - '0';
if (i < s2.length())t += s2[i] - '0';
int d = t % 10;
s.insert( 0, intstr(d));
c = t / 10;
}
return s;
}
//大数相减
string subtra(string &s1, string &s2) {
string s = "";
string flag;
removePrezero(s1); //去掉前置0
removePrezero(s2);
int len1 = s1.length();
int len2 = s2.length();
int len = len1>len2 ? len1 : len2;
if (len1 < len2)flag = "-";
else if (len1 > len2)flag = "+";
else {
int i;
for (i = 0; i < len1; i++) {
if (s1.at(i) > s2.at(i)) {
flag = "+"; break;
}
else if (s1.at(i) < s2.at(i)) {
flag = "-"; break;
}
}
if (i == len1)s == "0";
}
int* num = ( int*) malloc( sizeof( int)*len);
reverse(s1.begin(), s1.end()); //方便相减
reverse(s2.begin(), s2.end());
int c = 0;
for ( int j = 0; j < len; j++) {
int n1 = j < len1 ? s1[j] - '0' : 0;
int n2 = j < len2 ? s2[j] - '0' : 0;
if (flag == "+")num[c++] = n1 - n2;
else num[c++] = n2 - n1;
}
for ( int j = 0; j < c; j++) {
if (num[j] < 0) {
num[j] += 10; num[j + 1] -= 1;
}
}
c--;
while (num[c] == 0)c--;
for ( int j = 0; j <=c; j++){
s.insert( 0, intstr(num[j]));
}
if (flag == "-") return flag + s;
else return s;
}
//增加前置0
void addPrezero(string &s, int L) {
for ( int i = 0; i < L; i++)
s = s.insert( 0, "0");
}
//增加后置0，大数*10^n 相当于在数的后面加n个0
string addLastzero(string s, int L) {
string s1 = s;
for ( int i = 0; i < L; i++)
s1 += "0";
return s1;
}
//大数相乘
string multi(string &s1, string &s2) {
bool flag1 = false, flag2 = false;
string sign; //作用是判断结果是正数还是负数
if (s1.at( 0) == '-') {
flag1 = true; s1 = s1.substr( 1);
}
if (s1.at( 0) == '-') {
flag2 = true; s2 = s2.substr( 1);
}
if (flag1&&flag2)sign = "+";
else if (flag1 || flag2) sign = "-";
else sign = "+";
int L = 4; //将数前置若干个0，使其长度为2^n的倍数
if (s1.length() > 2 || s2.length() > 2) {
if (s1.length() >= s2.length()) {
while (L < s1.length())L *= 2;
if (L != s1.length())
addPrezero(s1, L - s1.length());
addPrezero(s2, L - s2.length());
} else {
while (L < s2.length())L*= 2;
if (L != s2.length())
addPrezero(s2, L - s2.length());
addPrezero(s1, L - s1.length());
}
}
if (s1.length() == 1)addPrezero(s1, 1);
if (s2.length() == 1)addPrezero(s2, 1);

int n = s1.length();
string result, a0, a1, b0, b1;
if (n > 1) {
a1 = s1.substr( 0, n / 2);
a0 = s1.substr(n / 2, n);
b1 = s2.substr( 0, n / 2);
b0 = s2.substr(n / 2, n);
}
if (n == 2) {
int x1 = strint(a1);
int x2 = strint(a0);
int y1 = strint(b1);
int y2 = strint(b0);
int num = (x1 * 10 + x2)*(y1 * 10 + y2);
result = intstr(num);
} else {
string c2 = multi(a1, b1);
string c0 = multi(a0, b0);
string temp1 = add(a0, a1);
string temp2 = add(b1, b0);
string temp3 = add(c2, c0);
string temp_c1 = multi(temp1, temp2);
string c1 = subtra(temp_c1, temp3);
string s1 = addLastzero(c1, n / 2);
string s2 =addLastzero(c2, n);
result = add(add(s1, s2), c0);
}
if (sign == "-")result.insert( 0, sign);
return result;
}
int main() {
//text
string s1 = "-1234999999999999999999999", s2 = "12348976543";
cout << multi(s1, s2);
}
  
  

总结：算法本身不难，主要是程序有太多需要注意的细节。对代码有疑问或者有些部分不理解的朋友，欢迎留言。

对了，程序的主要思想参考于点击打开链接，强烈建议大家可以去看一下。