47
import java.util.*;
public class Main {
public static void main(String[] args){
Scanner in=new Scanner(System.in);
String num1=in.nextBigDecimal().toString();
String num2=in.nextBigDecimal().toString();
int[] ret=new int[num1.length()+num2.length()];
for(int i=num1.length()-1;i>=0;i--){
int x=num1.charAt(i)-'0';
for(int j=num2.length()-1;j>=0;j--){
int y=num2.charAt(j)-'0';
ret[i+j]+=(ret[i+j+1]+x*y)/10;
ret[i+j+1]=(ret[i+j+1]+x*y)%10;
}
}
String s="";
for(int i=0;i
if(i==0&&ret[i]==0) continue;
s+=ret[i];
}
System.out.println(s);
}
}
发表于 2017-08-07 01:55:28
回复(19)
24
//AC代码:
#include
#include
#include
using namespace std;
const int L=11000;
string mul(string,string);
int main(){
string x,y;
while(cin>>x>>y)
cout<
}
string mul(string a,string b) {
string s;
int na[L],nb[L],nc[L],La=a.size(),Lb=b.size(),i,j;
fill(na,na+L,0);fill(nb,nb+L,0);fill(nc,nc+L,0);
for(i=La-1;i>=0;i--) na[La-i]=a[i]-'0';
for(i=Lb-1;i>=0;i--) nb[Lb-i]=b[i]-'0';
for(i=1;i<=La;i++)
for(j=1;j<=Lb;j++)
nc[i+j-1]+=na[i]*nb[j];
for(i=1;i<=La+Lb;i++)
nc[i+1]+=nc[i]/10,nc[i]%=10;
if(nc[La+Lb]) s+=nc[La+Lb]+'0';
for(i=La+Lb-1;i>=1;i--)
s+=nc[i]+'0';
return s;
}
发表于 2017-08-06 10:43:40
回复(13)
23
import java.util.Scanner;
/** Created by linjian on 17/8/5.
*/
public class Main {
public static void main(String[] args) {Scanner scanner = new Scanner(System.in);
String num1 = scanner.next();
String num2 = scanner.next();
System.out.println(multiply(num1,num2)); }
public static String multiply(String num1, String num2) {num1 = new StringBuilder(num1).reverse().toString();
num2 = new StringBuilder(num2).reverse().toString();
// even 99 * 99 is < 10000, so maximaly 4 digits
int[] d = new int[num1.length() + num2.length()];
for (int i = 0; i < num1.length(); i++) {
int a = num1.charAt(i) - '0';
for (int j = 0; j < num2.length(); j++) {
int b = num2.charAt(j) - '0';
d[i + j] += a * b;
}
}
StringBuilder sb = new StringBuilder();
for (int i = 0; i < d.length; i++) {
int digit = d[i] % 10;
int carry = d[i] / 10;
sb.insert(0, digit);
if (i < d.length - 1)
d[i + 1] += carry;
}
//trim starting zeros
while (sb.length() > 0 && sb.charAt(0) == '0') {
sb.deleteCharAt(0);
}
return sb.length() == 0 ? "0" : sb.toString(); }
}
发表于 2017-08-05 19:38:16
回复(10)
23
a,b = map(int,raw_input().split())
print str(a*b)
这竟然也过了，有点慌。。。
发表于 2017-08-07 11:04:11
回复(15)
6
import sys
s=sys.stdin.readline().strip().split()
def karatsuba_mul(num1, num2):
#karatsuba算法
if len(str(num1))==1 or len(str(num2))==1:
return num1*num2
n=max(len(str(num1)),len(str(num2)))
half=n//2
a=num1//10**half
b=num1%10**half
c=num2//10**half
d=num2%10**half
ac=karatsuba_mul(a,c) #计算a*c
bd=karatsuba_mul(b,d) #计算b*d
abcd=karatsuba_mul(a+b,c+d) #计算(a+b)*(c+d)
adbc=abcd-ac-bd
return ac*10**(2*half)+adbc*10**half+bd
print(str(karatsuba_mul(int(s[0]), int(s[1]))))
编辑于 2019-03-18 19:49:42
回复(2)
9
python a, b = map(int, input().split())
print(a * b)
发表于 2019-02-24 15:29:43
回复(3)
4
//写的不太好
#include 
#include 
using namespace std;
string addOfString(string s1, string s2)
{
string res = "";
if (s1.size()
s1 = string(s2.size() - s1.size(), '0') + s1;
else
s2 = string(s1.size() - s2.size(), '0') + s2;
int carry = 0;
for (int i = s1.size() - 1; i >= 0; --i)
{
char temp = (s1[i] - '0' + s2[i] - '0' + carry) % 10 + '0';
res = temp + res;
carry = (s1[i] - '0' + s2[i] - '0' + carry) / 10;
}
if (carry)    res = '1' + res;
return res;
}
string multiplyOfString(string s1, string s2)
{
string res = "";
string shorter = s1.size() < s2.size() ? s1 : s2;
string longer = s1.size() < s2.size() ? s2 : s1;
for (int i = shorter.size()-1; i >= 0; --i)
{
string temp = "";
for (int j = 0; j < shorter[i] - '0'; ++j)
temp = addOfString(temp, longer);
res = addOfString(temp + string(shorter.size() - i - 1, '0'), res);
}
return res;
}
int main()
{
string s1, s2;
cin >> s1 >> s2;
cout << multiplyOfString(s1, s2) << endl;
}
//看大家的评论发现可以直接乘不用加法，因此又写了一个方法二：
#include 
#include 
using namespace std;
string multiplyOfString(string s1, string s2)
{
string res(s1.size()+s2.size(),'0');
string shorter = s1.size() < s2.size() ? s1 : s2;
string longer = s1.size() < s2.size() ? s2 : s1;
for (int i = shorter.size() - 1; i >= 0; --i)
{
int carry = 0;//进位
for (int j = longer.size() - 1; j >= 0; --j)
{
int temp = (longer[j] - '0')*(shorter[i] - '0')+ carry + res[i+j+1] - '0';
res[i + j + 1] = temp % 10 + '0';
carry = temp / 10;
if (carry&&j == 0)//如果短串已经到最左边(j==0)并且carry是不为0的，那么res要修改；
res[i] += carry;
}
}
//去掉前排的0
int i = 0;
while (res[i] == '0')//此处不要写i++;
res.erase(0, 1);
return res;
}
int main()
{
string s1, s2;
cin >> s1 >> s2;
cout << multiplyOfString(s1, s2) << endl;
}
编辑于 2019-03-25 12:59:13
回复(6)
4
import java.util.Scanner;
/**
* @author wylu
*/
public class Main {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
while (scanner.hasNext()) {
char[] s1 = scanner.next().toCharArray();
char[] s2 = scanner.next().toCharArray();
int len = s1.length + s2.length;
char[] res = new char[len];
for (int i = 0; i < len; i++) res[i] = '0';
for (int i = s1.length - 1; i >= 0; i--) {
int cur = s1.length - i - 1, multiplier = s1[i] - '0', carry = 0;
for (int j = s2.length - 1; j >= 0; j--) {
int product = (res[cur] - '0') + multiplier * (s2[j] - '0') + carry;
res[cur++] = (char) (product % 10 + '0');
carry = product / 10;
}
if (carry != 0) res[cur] = (char) (carry + '0');
}
int i = len - 1;
while (i >= 0 && res[i] == '0') i--; //忽略前导0
if (i < 0) {
System.out.println(0);
}else {
for (; i >= 0; i--) System.out.print(res[i]);
System.out.println();
}
}
}
}
发表于 2019-01-12 00:53:10
回复(0)
3
Python版：通过十进制的运算，并且注意到十进制中可能发生的连续进位问题(不然可能通过率很低)，欢迎大家指正问题！
时间：40ms 内存：3568K
#创建指定大小的一维空数组
def kong(num):
s=[]
for i in range(num):
s.append(0);
return s
#十进制计算方法
def jinzhi(a,b):
cz=a+b;
if cz>= 10:
return [cz-10,1]
else:
return [cz,0]
#十进制中有连续进位，设置递归进行运算
def jinzhi2(c,i,j):
m=c[i+j]+1;
if m!=10:
c[i + j]=m;
else:
c[i+j]=0;
jinzhi2(c,i,j-1);
#主程序
def bignum_multi(cc):
dd = cc.split(' ');
a = dd[0];
b = dd[1];
m = len(a);
n = len(b);
p = m + n;#两个数相乘的位数为m+n-1或者m+n
c = kong(p);
#主程序
for i in range(n - 1, -1, -1):
for j in range(m - 1, -1, -1):
k = int(b[i]) * int(a[j]);
k1 = k // 10;
k2=k - k1 * 10;
t = jinzhi(c[i+j+1],k2);
c[i + j +1]=t[0];
if t[1] == 1:
jinzhi2(c, i, j);
t = jinzhi(c[i + j], k1);
c[i + j] = t[0];
if t[1] == 1:
jinzhi2(c, i, j-1);
#数组转成字符串
mm = "";
if c[0] == 0:
for i in range(1, p):
mm = mm + str(int(c[i]));
else:
for i in range(p):
mm = mm + str(int(c[i]));
print(mm)
cc=input();
bignum_multi(cc);
发表于 2019-02-28 19:15:34
回复(0)
3
#include 
#include 
#include 
using namespace std;
int main()
{
string num1, num2;
string s = "";
cin >> num1 >> num2;
int len1 = num1.size();
int len2 = num2.size();
vector vTemp(len1 + len2 - 1);
int n1, n2;
for (int i = 0; i
{
n1 = num1[i] - '0';
for (int j = 0; j 
{
n2 = num2[j] - '0';
vTemp[i + j] += n1*n2;
}
}
int carry = 0;
int tmp;
for (int i = vTemp.size()-1; i >= 0; i--)
{
tmp = vTemp[i] + carry;
vTemp[i] = tmp % 10;
s = to_string(vTemp[i])+s;
carry = tmp / 10;
}
if(carry > 0)
s = to_string(carry)+s;
cout << s << endl;
return 0;
}
发表于 2018-08-05 16:31:52
回复(3)
3
#include 
#include 
int multi(const char* lhs, const char* rhs, char* result) {
if (lhs == NULL || rhs == NULL || result == NULL)
return 0;
int llen = strlen(lhs), rlen = strlen(rhs);
if (llen == 0 || rlen == 0)
return 0;
// 用rhs每一位去乘lhs,模拟手工算乘法的过程。
int length = llen + rlen - 1;
for (int j = rlen - 1; j >= 0; j--) {
int carry = 0;// 乘法进位
for (int i = llen - 1; i >= 0; i--) {
int temp = (rhs[j] - '0') * (lhs[i] - '0') + carry;
// 根据位偏移将本位的结果加到result中去并处理加法进位。(result为倒序)
int bit = llen - i - 1 + (rlen - j - 1);
if ((result[bit] += temp % 10) >= '0' + 10) {
result[bit + 1] += (result[bit] - '0') / 10;
result[bit] = (result[bit] - '0') % 10 + '0';
}
// 保留进位。
carry = temp / 10;
}
// 最高位有进位
if (carry > 0) {
result[llen + (rlen - j - 1)] += carry;
length++;
}
}
// 去掉多余的0
while (result[length - 1] == '0')
length--;
return length;
}
int main() {
char a[999];
char b[999];
scanf("%s%s", a, b);
char r[999999];
memset(r, '0', sizeof(r));
int length = multi(a, b, r);
while (--length >= 0) {
std::cout << r[length];
}
system("pause");
return 0;
}
发表于 2017-08-06 22:30:16
回复(0)
6
# coding=utf-8
def fun(l):
num0,num1 = l[0],l[1]
result = 0
# 还原乘法的过程，按位相乘再乘以位数权重
for i in range(len(num0)):
for j in range(len(num1)):
# 注意计算位数权重时，index越大位数越小
result += int(num0[i])*int(num1[j])*pow(10,len(num0)-1-i+len(num1)-1-j)
return (str(result))
if __name__=='__main__':
s = raw_input()
l = s.split(' ')
print(fun(l))
发表于 2019-01-15 16:03:15
回复(1)
2
模拟手算过程：
import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;
public class Main {
private static String add(String n1, String n2) {
StringBuilder res = new StringBuilder();
int p1 = n1.length() - 1;
int p2 = n2.length() - 1;
int carry = 0;
while (carry > 0 || p1 >= 0 || p2 >= 0) {
int value = carry + (p1 >= 0 ? n1.charAt(p1) - '0' : 0) + (p2 >= 0 ? n2.charAt(p2) - '0' : 0);
res.append(value % 10);
carry = value / 10;
if (p1 >= 0) p1--;
if (p2 >= 0) p2--;
}
return res.reverse().toString();
}
private static String singleMul(String n1, char n2) {
StringBuilder res = new StringBuilder();
int p = n1.length() - 1;
int carry = 0;
while (carry > 0 || p >= 0) {
int value = carry + (n2 - '0') * (p >= 0 ? n1.charAt(p) - '0' : 0);
res.append(value % 10);
carry = value / 10;
if (p >= 0) p--;
}
return res.reverse().toString();
}
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
String n1 = scanner.next();
String n2 = scanner.next();
if (n1.length() < n2.length()) {
String tmp = n1;
n1 = n2;
n2 = tmp;
}
List sums = new ArrayList<>();
for (int i = n2.length() - 1; i >= 0; i--) {
sums.add(singleMul(n1, n2.charAt(i)) + new String(new char[n2.length() - 1 - i]).replace('\0', '0'));
}
String res = "0";
for (String s : sums) {
res = add(res, s);
}
System.out.println(res);
}
}
运行时间：74ms
占用内存：11272k
发表于 2019-03-13 00:33:42
回复(0)
2
N,M = input().strip().split()
print(eval（N) * eval（M))
一个用python答题者的心里过程：
为了显示我的能力，我表示我不用库函数都能写出来！
对着题思考十秒钟...
呵呵，这个库函数真好用
发表于 2018-01-25 16:48:58
回复(0)
1
直接模拟乘法的竖式计算过程，不过要注意计算加法的时候数字排列是阶梯状的，需要在数的后面补0。
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.IOException;
import java.util.Stack;
public class Main {
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String[] strNum = br.readLine().trim().split(" ");
System.out.println(multiply(strNum[0].toCharArray(), strNum[1].toCharArray()));
}
// 模拟乘法竖式
private static String multiply(char[] num1, char[] num2) {
if(num1.length 
char[] temp = num1;
num1 = num2;
num2 = temp;
}
Stack stack = new Stack<>();
for(int j = num2.length - 1; j >= 0; j--){
StringBuilder sb = new StringBuilder();
// 之后做竖式加法要补0
for(int k = 0; k 
int carry = 0;
for(int i = num1.length - 1; i >= 0; i--){
int num = (num1[i] - '0')*(num2[j] - '0') + carry;
carry = num / 10;
sb.append(num % 10);
}
if(carry > 0) sb.append(carry);
if(stack.isEmpty())
stack.push(sb.reverse().toString());
else{
String temp = stack.pop();
stack.push(add(temp, sb.reverse().toString()));
}
}
return stack.pop();
}
private static String add(String num1, String num2) {
if(num1.length() 
String temp = num1;
num1 = num2;
num2 = temp;
}
int len1 = num1.length();
int len2 = num2.length();
int i = len1 - 1, j = len2 - 1;
int carry = 0;
StringBuilder sb = new StringBuilder();
while(i >= 0 && j >= 0){
int num = (num1.charAt(i) - '0') + (num2.charAt(j) - '0') + carry;
carry = num / 10;
sb.append(num % 10);
i--;
j--;
}
while(i >= 0){
int num = (num1.charAt(i) - '0') + carry;
carry = num / 10;
sb.append(num % 10);
i--;
}
if(carry > 0) sb.append(carry);
return sb.reverse().toString();
}
}
发表于 2021-02-01 14:57:29
回复(0)
1
#include 
using namespace std;
int main(){
string s1, s2;
cin>>s1>>s2;
int n=s1.size(), m=s2.size(), k;
string r(n+m,'0');
for(int i=n-1;i>=0;i--){
int c = 0;
for(int j=m-1;j>=0;j--){
int t = (r[i+j+1]-'0') + (s1[i]-'0')*(s2[j]-'0') + c;
r[i+j+1] = t%10 + '0';
c = t/10;
}
r[i] += c;
}
for(k=0;k
if(r[k]!='0')
break;
r = r.substr(k);
cout<
return 0;
}
发表于 2020-11-10 01:10:00
回复(0)
1
from functools import reduce
def multiply(num1, num2):
n1, n2 = len(num1), len(num2)
if n1 
res = []
for j in range(n2 - 1, -1, -1):
tmpRes = ''
carry = 0
for i in range(n1 - 1, -1, -1):
tmp = int(num1[i]) * int(num2[j]) + carry
tmpRes = str(tmp % 10) + tmpRes
carry = tmp // 10
tmpRes = str(carry) + tmpRes if carry != 0 else tmpRes
if int(tmpRes) != 0:
tmpRes += (n2 - j - 1) * '0'
else:
tmpRes = '0'
res.append(tmpRes)
return reduce(addStrings, res) if len(res) > 1 else res[0]
def addStrings(num1, num2):
i, j, res = len(num1) - 1, len(num2) - 1, ''
carry = 0
while i >= 0&nbs***bsp;j >= 0&nbs***bsp;carry:
c1 = int(num1[i]) if i >= 0 else 0
c2 = int(num2[j]) if j >= 0 else 0
tmp = c1 + c2 + carry
carry = tmp // 10
res = str(tmp % 10) + res
i, j = i - 1, j - 1
return res
a, b = input().split()
print(multiply(a, b))
发表于 2020-07-15 16:24:50
回复(0)
1
import java.math.BigDecimal;
import java.util.*;
public class Main {
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner cin=new Scanner (System.in);
String s1=cin.next();
String s2=cin.next();
BigDecimal b1=new BigDecimal(s1);
BigDecimal b2=new BigDecimal(s2);
BigDecimal b3=b1.multiply(b2);
String s3=String.valueOf(b3);
System.out.print(s3);
}
}
发表于 2020-01-10 12:32:45
回复(0)
1
#include 
#include 
#include 
#include 
using namespace std;
void output(vector v)
{
for (auto ele : v)
{
cout <
}
putchar(10);
}
vector mul(string x, string y)
{
vector res(x.length() + y.length(), 0);
reverse(x.begin(), x.end());
reverse(y.begin(), y.end());
//先乘，不用考虑进位，int范围21亿，10以内的乘法，可以存2千万次
for (int i = 0; i 
{
for (int j = 0; j 
{
//和x的第一位数乘，结果从res[0]开始保存，和x的第二位数乘，结果从res[1]开始保存，以此类推
res[i + j] += (x[i] - '0') * (y[j] - '0');
}
}
//最后一次性进位
for (int i = 0; i 
{
res[i + 1] += res[i] / 10;
res[i] %= 10;
}
//3 * 3 = 9，但是申请了2个int的vector，最后一个int没有进位，pop掉
while (res.back() == 0)
{
res.pop_back();
}
//逆置就是结果了
reverse(res.begin(), res.end());
return res;
}
int main()
{
string x, y;
while (cin >> x >> y)
{
output(mul(x, y));
}
return 0;
}

1
a, b = map(int, input().split())
print(str(a*b))