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：
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#include
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#include
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#include
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#include
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using
namespace
std;
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//string转换为int
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int strint(string s) {
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int num;
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stringstream ss(s);
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ss >> num;
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return num;
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}
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//int转换为string
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string intstr(int num) {
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string s;
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stringstream ss;
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ss << num;
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ss >> s;
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return s;
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}
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//前置0 ，因为大数的长度必须要为2^n的倍数，才能划分
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void removePrezero(string &s) {
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if (s.length() ==
1)
return;
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int k =
0;
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for (
int i =
0; i < s.length(); i++) {
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if (s[i] ==
'0')k++;
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else
break;
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}
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if (k == s.length())s =
"0";
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else s = s.substr(k);
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}
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//大数相加
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string add(string s1, string s2) {
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string s =
"";
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removePrezero(s1);
//去掉前置0
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removePrezero(s2);
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reverse(s1.begin(), s1.end());
//先把字符串颠倒，方便相加
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reverse(s2.begin(), s2.end());
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int c =
0;
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for (
int i =
0;c|| i < max(s1.length(), s2.length()); i++) {
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int t = c;
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if (i < s1.length())t += s1[i] -
'0';
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if (i < s2.length())t += s2[i] -
'0';
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int d = t %
10;
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s.insert(
0, intstr(d));
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c = t /
10;
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}
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return s;
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}
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//大数相减
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string subtra(string &s1, string &s2) {
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string s =
"";
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string flag;
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removePrezero(s1);
//去掉前置0
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removePrezero(s2);
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int len1 = s1.length();
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int len2 = s2.length();
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int len = len1>len2 ? len1 : len2;
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if (len1 < len2)flag =
"-";
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else
if (len1 > len2)flag =
"+";
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else {
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int i;
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for (i =
0; i < len1; i++) {
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if (s1.at(i) > s2.at(i)) {
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flag =
"+";
break;
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}
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else
if (s1.at(i) < s2.at(i)) {
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flag =
"-";
break;
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}
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}
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if (i == len1)s ==
"0";
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}
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int* num = (
int*)
malloc(
sizeof(
int)*len);
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reverse(s1.begin(), s1.end());
//方便相减
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reverse(s2.begin(), s2.end());
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int c =
0;
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for (
int j =
0; j < len; j++) {
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int n1 = j < len1 ? s1[j] -
'0' :
0;
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int n2 = j < len2 ? s2[j] -
'0' :
0;
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if (flag ==
"+")num[c++] = n1 - n2;
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else num[c++] = n2 - n1;
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}
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for (
int j =
0; j < c; j++) {
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if (num[j] <
0) {
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num[j] +=
10; num[j +
1] -=
1;
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}
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}
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c--;
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while (num[c] ==
0)c--;
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for (
int j =
0; j <=c; j++){
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s.insert(
0, intstr(num[j]));
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}
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if (flag ==
"-")
return flag + s;
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else
return s;
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}
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//增加前置0
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void addPrezero(string &s, int L) {
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for (
int i =
0; i < L; i++)
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s = s.insert(
0,
"0");
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}
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//增加后置0，大数*10^n 相当于在数的后面加n个0
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string addLastzero(string s, int L) {
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string s1 = s;
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for (
int i =
0; i < L; i++)
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s1 +=
"0";
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return s1;
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}
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//大数相乘
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string multi(string &s1, string &s2) {
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bool flag1 =
false, flag2 =
false;
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string sign;
//作用是判断结果是正数还是负数
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if (s1.at(
0) ==
'-') {
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flag1 =
true; s1 = s1.substr(
1);
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}
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if (s1.at(
0) ==
'-') {
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flag2 =
true; s2 = s2.substr(
1);
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}
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if (flag1&&flag2)sign =
"+";
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else
if (flag1 || flag2) sign =
"-";
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else sign =
"+";
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int L =
4;
//将数前置若干个0，使其长度为2^n的倍数
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if (s1.length() >
2 || s2.length() >
2) {
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if (s1.length() >= s2.length()) {
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while (L < s1.length())L *=
2;
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if (L != s1.length())
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addPrezero(s1, L - s1.length());
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addPrezero(s2, L - s2.length());
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}
else {
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while (L < s2.length())L*=
2;
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if (L != s2.length())
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addPrezero(s2, L - s2.length());
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addPrezero(s1, L - s1.length());
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}
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}
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if (s1.length() ==
1)addPrezero(s1,
1);
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if (s2.length() ==
1)addPrezero(s2,
1);
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int n = s1.length();
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string result, a0, a1, b0, b1;
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if (n >
1) {
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a1 = s1.substr(
0, n /
2);
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a0 = s1.substr(n /
2, n);
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b1 = s2.substr(
0, n /
2);
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b0 = s2.substr(n /
2, n);
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}
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if (n ==
2) {
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int x1 = strint(a1);
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int x2 = strint(a0);
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int y1 = strint(b1);
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int y2 = strint(b0);
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int num = (x1 *
10 + x2)*(y1 *
10 + y2);
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result = intstr(num);
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}
else {
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string c2 = multi(a1, b1);
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string c0 = multi(a0, b0);
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string temp1 = add(a0, a1);
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string temp2 = add(b1, b0);
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string temp3 = add(c2, c0);
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string temp_c1 = multi(temp1, temp2);
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string c1 = subtra(temp_c1, temp3);
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string s1 = addLastzero(c1, n /
2);
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string s2 =addLastzero(c2, n);
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result = add(add(s1, s2), c0);
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}
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if (sign ==
"-")result.insert(
0, sign);
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return result;
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}
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int main() {
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//text
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string s1 =
"-1234999999999999999999999", s2 =
"12348976543";
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cout << multi(s1, s2);
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}
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。