各位大佬们，求调

P1303 A*B Problem

Jerry555 @ 2024-02-17 17:22:55

#include <iostream>
#include <string>

using namespace std;

class Basic_tools {
public:
    std::string removeLeadingZeros(const std::string& str) {
        size_t pos = str.find_first_not_of('0');
        if (pos != std::string::npos) {
            return str.substr(pos);
        }
        return "0";
    }
}JTools;

class BigNumber {
public:
    std::string number;
    int len;
    bool empty = true;
    BigNumber() {
        number="";
        len=0;
    }
    BigNumber(std::string NewNum) {
        number = NewNum;
        len = NewNum.size();
        empty = false;
    }
    void InitIn() {
        std::cin >> number;
        len = number.size();
        empty = false;
    }
    void Output() {
        std::cout << number;
    }
    void Delete() {
        empty = true;
        number = "";
        len = 0;
    }
};
BigNumber operator*(BigNumber a, BigNumber b) {
    int A[5010]={0}, B[5010]={0},D[5010]={0};
    //BigNumber D;
    for (int i = a.len-1; i >= 0; i--)A[a.len - i] = a.number[i] - '0';
    for (int i = b.len-1; i >= 0; i--)B[b.len - i] = b.number[i] - '0';
    for (int i = 1; i <= a.len; i++)for (int j = 1; j <= b.len; j++) D[i + j - 1] += A[i] * B[j];
    int len=a.len+b.len;
    for (int i = 1; i <= len; i++) {
        D[i + 1] += D[i] / 10;
        D[i] %= 10;
    }
    string Ds;
    for(int i=max(1,len);i>=1;i--){
        Ds.insert(0,to_string(D[i]+'0'));
    }
    JTools.removeLeadingZeros(Ds);
    return BigNumber(Ds);
}
int main(){
    BigNumber AB,BB;
    AB.InitIn();
    BB.InitIn();
    BigNumber CB;
    CB = AB * BB;
    CB.Output();
}

by wuzhiyuan123 @ 2024-02-21 20:58:08

dalao啊，你这个代码我看着都不像这题的，我都看不懂了，有点复杂，我给你我的代码参考一下

by wuzhiyuan123 @ 2024-02-21 21:07:41

#include <iostream>
#include <cstdio>
#include <string>
using namespace std;
string a1,b1;
int a[1000010],b[1000010],c[1000010];
int main()
{
    //freopen("input.txt", "r", stdin);//这行代码在提交的时候记得删掉，如果你不知道这行代码的意思可以删掉 
    cin >> a1 >> b1;
    int len1 =0;
    //不想算了，特判一下 
    if((a1.length()==1&&a1[1]==0)||(b1.length()==1&&b1[1]==0)){
        cout << 0;
        return 0;
    }
    //输入，吧正着的倒过来 
    for(int i=a1.length()-1, j=1;i>=0;i--, j++){
        a[j] = a1[i] - '0';
    } 
    for(int i=b1.length()-1, j=1;i>=0;i--, j++){
        b[j] = b1[i] - '0';
    } 
    //开始做乘法，把a[i]*b[j]的结果放的c[i+j-1]，至于为什么这么放画个表格就知道了 
    for(int i=1;i<=a1.length();i++){
        for(int j=1;j<=b1.length();j++){
            c[i+j-1]+=a[i]*b[j];
            len1=max(i+j-1, len1);
        }
    }
    //进位 
    int len2=len1;
    for(int i=1;i<=len2;i++){
        if(c[i]>=10){
            int change = c[i]/10;
            c[i]=c[i]%10;
            c[i+1]+=change;
            if(i==len2) len1++;
        } 
    }
    //倒着输出得到正的数 
    for(int i=len1;i>0;i--){
        cout << c[i];
    }
    return 0;
}