大整数乘法
问题描述
求两个不超过200位的非负整数的积。
输入形式
有两行，每行是一个不超过200位的非负整数，没有多余的前导0。
输出形式
一行，即相乘后的结果。结果里不能有多余的前导0，即如果结果是342，那么就不能输出为0342。
样例输入
1234567890
9876543210
样例输出
12193263111263526900
解题思路：
1 :采用了数组的形式进行计算。将输入的值以字符的形式存到数组中取，再转化成整数形式相乘。
2. 根据课本的描述采用分治的方法求解。主要部分是递归的不同种情况。当n=2时直接求解，当n>2，则对其进行分治。按照公式：A*B=a1*b1*10^n+[(a1+a0)*(b0+b1)-a1*a0-b1*b0]*10^n/2+a0*b0求解
#include<iostream>
#include<cstring>
using namespace std;
int main(){
char A[200],B[200];
int X[200],Y[200];
int Z[401]={0};
cin>>A;
cin>>B;
int lenA = strlen(A);
int lenB = strlen(B);
int i,j;
for(i=lenA-1,j=0;i>=0;i--)
{
X[j]=A[i]-'0';
j++;
}
for(i=lenB-1,j=0;i>=0;i--)
{
Y[j]=B[i]-'0';
j++;
}
for(i=0;i<lenA;i++)
{
for(j=0;j<lenB;j++)
Z[i+j]=Z[i+j]+X[i]*Y[j];
}
for(i=0;i<400;i++)
{
if(Z[i]>=10)
{
Z[i+1]=Z[i+1]+Z[i]/10;
Z[i]=Z[i]%10;
}
}
for(i=400;i>0;i--)
{
if(Z[i]==0) continue;
else break;
}
for(;i>=0;i--)
cout<<Z[i];
return 0;
}
第二种：
#include<iostream>
#include<string>
#include<algorithm>
#include<sstream>
using namespace std;
//string转换为int
int str_to_int(string s) {
int num;
stringstream ss(s);
ss >> num;
return num;
}
//int转换为string
string int_to_str(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
//去掉前置零，substr（）复制子字符串
void removePrezero(string &s) {
int i = 0;
while (i < s.length() && s[i] == '0') //过滤前置零
i++;
if (i < s.length()) s = s.substr(i); //复制剩余子字符串
else s = "0";
}
//大数相加
string add(string s1, string s2) {
string s = "";
//去掉前置0
removePrezero(s1);
removePrezero(s2);
//先把字符串颠倒，方便相加
reverse(s1.begin(), s1.end());
reverse(s2.begin(), s2.end());
int c = 0; //记录进位值
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 = char(d + '0') + s;
c = t / 10;
}
return s;
}
//大数相减
string sub(string &s1, string &s2) {
string s = "";
string flag;
//去掉前置0
removePrezero(s1);
removePrezero(s2);
//求长比较，确定正负
int len1 = s1.length();
int len2 = s2.length();
int len = len1>len2 ? len1 : len2;
if (len1 < len2)flag = "-";
else if (len1 > len2)flag = "+";
else { //s.at(n) 返回s中下标为n的元素
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());
reverse(s2.begin(), s2.end());
int c = 0; //记录结果位数
for (int i = 0; i < len; i++) {
int n1 = i < len1 ? s1[i] - '0' : 0;
int n2 = i < len2 ? s2[i] - '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--;
//去除可能存在的前置零
while (num[c] == 0)c--;
for (int j = 0; j <=c; j++){
s = char(num[j] + '0') + s ;
}
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 mult(string &s1, string &s2) {
bool flag1 = false, flag2 = false;
string sign; //记录结果正负
if (s1.at(0) == '-') {
flag1 = true; s1 = s1.substr(1);
}
if (s2.at(0) == '-') {
flag2 = true; s2 = s2.substr(1);
}
if (flag1 || flag2) sign = "-";
else sign = "+";
int L = 4;
//前置若干个0，使长度为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);
//进行运算
int n = s1.length();
string result, a0, a1, b0, b1;
if (n > 1) { //大于1取n/2
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) { //n==2 计算
int x1 = str_to_int(a1);
int x2 = str_to_int(a0);
int y1 = str_to_int(b1);
int y2 = str_to_int(b0);
int num = (x1 * 10 + x2)*(y1 * 10 + y2);
result = int_to_str(num);
}else { // n!=2 采用公式求解
//A*B=a1*b1*10^n+[(a1+a0)*(b0+b1)-a1*a0-b1*b0]*10^n/2+a0*b0
string c2 = mult(a1, b1);
string c0 = mult(a0, b0);
string temp1 = add(a0, a1);
string temp2 = add(b1, b0);
string temp3 = add(c2, c0);
string temp_c1 = mult(temp1, temp2);
string c1 = sub(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() {
string s1 ,s2;
cin>>s1>>s2;
cout << mult(s1, s2);
}