分治（Divide and Conquer）算法：问题可以分解为子问题，每一个问题是可以独立的解决的，从子问题的解可以构建原问题。
Divide：中间分、随机分、奇偶分等，将问题分解成独立的子问题
Conquer：子问题的解可以单独解决，从子问题的解构建原问题终究的解
Combine：每步将子问题产生的解进行合并得到终究的解，合并的复杂度影响终究的算法时间复杂度
Karatsuba算法是在普通乘法算法的基础上进行的提升，使得终究的复杂度从O(n^2)变成了O(n^1.585)，基本思想是将原问题的范围每次减小1般，并且每次解决3个子问题：
X = Xl*2n/2 + Xr [Xl 左边n/2位数 Xr 右边n/2位数]
Y = Yl*2n/2 + Yr [Yl 左边n/2位数 Yr 右边n/2位数]
XY = (Xl*2n/2 + Xr)(Yl*2n/2 + Yr)
= 2n XlYl + 2n/2(XlYr + XrYl) + XrYr
XY = 2n XlYl + 2n/2 * [(Xl + Xr)(Yl + Yr) - XlYl - XrYr] + XrYr
XY = 22ceil(n/2) XlYl + 2ceil(n/2) * [(Xl + Xr)(Yl + Yr) - XlYl - XrYr] + XrYr
从
从而得到终究的算法时间复杂度为T(n) = 3T(n/2) + O(n)，得到T(n) = O(n^1.585)。算法的伪代码以下：
karatsuba(num1, num2)
if (num1 < 10) or (num2 < 10)
return num1*num2
/* calculates the size of the numbers */
m = max(size_base10(num1), size_base10(num2))
m2 = m/2
/* split the digit sequences about the middle */
high1, low1 = split_at(num1, m2)
high2, low2 = split_at(num2, m2)
/* 3 calls made to numbers approximately half the size */
z0 = karatsuba(low1,low2)
z1 = karatsuba((low1+high1),(low2+high2))
z2 = karatsuba(high1,high2)
return (z2*10^(2*m2))+((z1-z2-z0)*10^(m2))+(z0)
下面是使用C++具体实现的进程，如果直接使用整数类型实现，可能会产生溢出，因此使用输入的字符串表示，实际运算的进程将字符串转换为数组进行加、减、乘操作。先看终究的算法实现：
string Multiplicate(string x, string y)
{
int len = GetSameSize(x, y);
if (len == 0) return 0;
if (len == 1) return MultiplyString(x, y);
int p = len % 2 == 0 ? len / 2 : len / 2 + 1;
string Xh = x.substr(0, len / 2);
string Yh = y.substr(0, len / 2);
string Xl = x.substr(len / 2);
string Yl = y.substr(len / 2);
string P1 = Multiplicate(Xh, Yh);
string P2 = Multiplicate(Xl, Yl);
string P3 = Multiplicate(AddString(Xh, Xl), AddString(Yh, Yl));
return
AddString(
AddString(
MultiplyPower(P1, 2 * p),
MultiplyPower(MinusString(MinusString(P3, P1), P2), p)
), P2
);
}
上述就是依照伪代码进行实现，但是使用了字符串的数字运算操作，包括字符串与数组的转换，数组加、减、乘，具体实现以下：
void StringToArray(string a, int *arr)
{
int n = a.size();
for(int i = 0; i < n; i++)
arr[n - i - 1] = a.at(i) - '0';
}
void ArrayToString(int *arr, int len, string & a)
{
for(int i = 0; i < len; i++)
a += '0' + arr[len - i - 1];
}
string DelPreZero(string a)
{
int zeros = 0;
for (int i = 0; i < a.size(); i++)
if (a.at(i) == '0') zeros++;
else break;
if (zeros == a.size()) return "0";
return a.substr(zeros);
}
void MultiplyArray(int a[], int la, int b[], int lb, int *arr)
{
int i;
for (i = 0; i < la; i++)
for (int j = 0; j < lb; j++)
arr[i + j] += a[i] * b[j];
for (i = 0; i < la + lb - 1; i++)
{
arr[i + 1] += arr[i] / 10;
arr[i] = arr[i] % 10;
}
}
void AddArray(int a[], int la, int b[], int lb, int *arr)
{
int i;
int len = la > lb ? lb : la;
for (i = 0; i < len; i++)
arr[i] += a[i] + b[i];
if (la > lb)
{
for (i = lb; i < la; i++)
arr[i] = a[i];
for (i = 0; i < la; i++)
{
arr[i + 1] += arr[i] / 10;
arr[i] = arr[i] % 10;
}
}
else
{
for (i = la; i < lb; i++)
arr[i] = b[i];
for (i = 0; i < lb; i++)
{
arr[i + 1] += arr[i] / 10;
arr[i] = arr[i] % 10;
}
}
}
void MinusArray(int a[], int la, int b[], int lb, int *arr) //a must be bigger than b
{
int i;
for (i = 0; i < lb; i++)
arr[i] = a[i] - b[i];
for (i = lb; i < la; i++)
arr[i] = a[i];
for (i = 0; i < la - 1; i++)
{
if (arr[i] < 0)
{
arr[i + 1]--;
arr[i] = 10 + arr[i];
}
}
}
string MultiplyString(string a, string b)
{
int m = a.size(), n = b.size();
int *arrA = new int[m];
int *arrB = new int[n];
StringToArray(a, arrA);
StringToArray(b, arrB);
int *arrC = new int[m + n];
for(int i = 0; i < n + m; i++) arrC[i] = 0;
string rst;
MultiplyArray(arrA, m, arrB, n, arrC);
ArrayToString(arrC, m + n, rst);
delete []arrA;
delete []arrB;
delete []arrC;
return DelPreZero(rst);
}
string AddString(string a, string b)
{
int m = a.size(), n = b.size();
int *arrA = new int[m];
int *arrB = new int[n];
StringToArray(a, arrA);
StringToArray(b, arrB);
int i, len = m > n ? m : n;
int *arrC = new int[len + 1];
for(i = 0; i < len + 1; i++) arrC[i] = 0;
AddArray(arrA, m, arrB, n, arrC);
string rst;
ArrayToString(arrC, len + 1, rst);
delete []arrA;
delete []arrB;
delete []arrC;
return DelPreZero(rst);
}
string MultiplyPower(string a, int len)
{
for(int i = 0; i < len; i++)
a += '0';
return DelPreZero(a);
}
string MinusString(string a, string b)
{
int m = a.size(), n = b.size();
int *arrA = new int[m];
int *arrB = new int[n];
StringToArray(a, arrA);
StringToArray(b, arrB);
string rst;
int i, len = m > n ? m : n;
int *arrC = new int[len];
for(i = 0; i < len; i++) arrC[i] = 0;
MinusArray(arrA, m, arrB, n, arrC);
ArrayToString(arrC, len, rst);
delete []arrA;
delete []arrB;
delete []arrC;
return DelPreZero(rst);
}
主要是触及到字符串与数组的转换中字符串在数字中是逆序的，进行数组运算时方便，同时对数组间的减法，只支持a 大于b的减法，如果是a 小于b可以用b减去a后再取反便可。还有就是对数组的动态空间申请后，需要及时释放。
参考：
1.http://www.geeksforgeeks.org/divide-and-conquer-set⑵-karatsuba-algorithm-for-fast-multiplication/
2.http://en.wikipedia.org/wiki/Karatsuba_algorithm#Pseudo_Code_Implementation