C++高精度大数运算

一、大数相加（两个正数）

1、最简单的，两个整形大数相加

练习：HDU1002

代码：

#include 
#include 
#include 
using namespace std;

// 两个整形大数相加
string Add(string &num1, string &num2){
    string ans = "";
    int l1 = num1.length()-1;
    int l2 = num2.length()-1;
    int tmp1 = 0, tmp2 = 0;
    while(l1 >= 0 && l2 >= 0){
        tmp2 = tmp1 + num1[l1] - '0' + num2[l2] - '0';
        tmp1 = tmp2 / 10;
        ans += char(tmp2 % 10 + '0');
        l1--, l2--;
    }
    while(l1 >= 0){
        tmp2 = tmp1 + num1[l1] - '0';
        tmp1 = tmp2 / 10;
        ans += char(tmp2 % 10 + '0');
        l1--;
    }
    while(l2 >= 0){
        tmp2 = tmp1 + num2[l2] - '0';
        tmp1 = tmp2 / 10;
        ans += char(tmp2 % 10 + '0');
        l2--;
    }
    while(tmp1){
        ans += char(tmp1 % 10 + '0');
        tmp1 /= 10;
    }
    reverse(ans.begin(), ans.end());
    return ans;
}

int main(){
    int t;
    string num1, num2;
    cin>>t;
    for(int i = 1; i <= t; i++){
        cin>>num1>>num2;
        string ans = Add(num1, num2);
        if(i > 1) cout<#include 
#include 
#include 
using namespace std;

int find_ch(string &str, char ch){
    for(int i = 0; i < str.length(); i++)
        if(str[i] == ch) return i;
    return -1;
}
string Add(string &num1, string &num2){
    string ans = "";
    int p1 = find_ch(num1, '.'), p2 = find_ch(num2, '.');
    string num11 = "", num21 = "";
    if(p1 != -1) num11 = num1.substr(p1+1, num1.length()-p1-1);
    if(p2 != -1) num21 = num2.substr(p2+1, num2.length()-p2-1);
    int l11 = num11.length()-1, l21 = num21.length()-1;
    while(l11 > l21){
        ans += num11[l11--];
    }
    while(l21 > l11){
        ans += num21[l21--];
    }
    int tmp1 = 0, tmp2 = 0;
    while(l11 >= 0){
        tmp2 = tmp1 + (num11[l11] - '0') + (num21[l21] - '0');
        tmp1 = tmp2 / 10;
        ans += char(tmp2 % 10 + '0');
        l11--, l21--;
    }
    ans += '.';
    string num12, num22;
    if(p1 == -1) num12 = num1;
    else num12 = num1.substr(0, p1);
    if(p2 == -1) num22 = num2;
    else num22 = num2.substr(0, p2);
    int l12 = num12.length()-1, l22 = num22.length()-1;
    while(l12 >= 0 && l22 >= 0){
        tmp2 = tmp1 + (num12[l12] - '0') + (num22[l22] - '0');
        tmp1 = tmp2 / 10;
        ans += char(tmp2 % 10 + '0');
        l12 --, l22 --;
    }
    while(l12 >= 0){
        tmp2 = tmp1 + (num12[l12] - '0');
        tmp1 = tmp2 / 10;
        ans += char(tmp2 % 10 + '0');
        l12 --;
    }
    while(l22 >= 0){
        tmp2 = tmp1 + (num22[l22] - '0');
        tmp1 = tmp2 / 10;
        ans += char(tmp2 % 10 + '0');
        l22 --;
    }
    while(tmp1){
        ans += char(tmp1 % 10 + '0');
        tmp1 /= 10;
    }
    reverse(ans.begin(), ans.end());
    int p = ans.length() - 1;
    // cout<<"* "<>num1>>num2){
        string ans = Add(num1, num2);
        cout<

二、大数相减（两个正数）

两个正整数相减

代码：

#include 
#include 
#include 
using namespace std;

string Sub(string &num1, string &num2){
    string ans = "";
    int l1 = num1.length() - 1;
    int l2 = num2.length() - 1;
    bool neg = false;
    if(l2 > l1 || num1 < num2){
        string tmp = num1;
        num1 = num2;
        num2 = tmp;
        neg = true;
        swap(l1, l2);
    }
    int tmp1 = 0, tmp2 = 0;
    while(l2 >= 0){
        tmp2 = (num1[l1] - num2[l2]) - tmp1;
        if(tmp2 < 0){
            tmp2 += 10;
            tmp1 = 1;
        }else{
            tmp1 = 0;
        }
        ans += char(tmp2 + '0');
        l1--, l2--;
    }
    while(l1 >= 0){
        tmp2 = (num1[l1] - '0') - tmp1;
        if(tmp2 < 0){
            tmp2 += 10;
            tmp1 = 1;
        }else{
            tmp1 = 0;
        }
        ans += char(tmp2 + '0');
        l1--;
    }
    int p = ans.length() - 1;
    while(ans[p] == '0' && p > 0) p --;
    ans = ans.substr(0, p+1);
    if(neg) ans += '-';
    reverse(ans.begin(), ans.end());
    return ans;
}

int main(){
    string num1, num2;
    while(cin>>num1>>num2){
        string ans = Sub(num1, num2);
        cout<

三、大数乘法（两个正数）

两个正整数相乘

练习：HDU1402

超时代码（直接模拟乘法运算，复杂度是O(nm)，这个题会超时）：

#include 
#include 
#include 
#include 
using namespace std;

const int N = 50005;
string Mul(string &num1, string &num2){
    int pre_zero = 0; // 处理前导0
    while(num1[pre_zero] == '0' && pre_zero < num1.length()-1) pre_zero ++;
    num1 = num1.substr(pre_zero, num1.length() - pre_zero);
    pre_zero = 0;
    while(num2[pre_zero] == '0' && pre_zero < num2.length()-1) pre_zero ++;
    num2 = num2.substr(pre_zero, num2.length()-pre_zero);
    string zero = "0";
    if(num1 == zero || num2 == zero) return zero;

    string ans = "";
    int l1 = num1.length() - 1;
    int l2 = num2.length() - 1;
    int tmp1 = 0, tmp2 = 0, start = 0;
    while(l1 >= 0){
        int p = start ++;
        for(int i = l2; i >= 0; i--){
            tmp1 = tmp1 + (num1[l1] - '0') * (num2[i] - '0');
            if(ans.length() > p){
                tmp2 = ans[p] - '0' + tmp1;
                ans[p] = char(tmp2 % 10 + '0');
                tmp1 = tmp2 / 10;
            }else{
                ans += char(tmp1 % 10 + '0');
                tmp1 /= 10;
            }
            p ++;
        }
        while(tmp1){
            ans += char(tmp1%10 + '0');
            tmp1 /= 10;
        }
        l1 --;
    }
    reverse(ans.begin(), ans.end());
    return ans;
}
int main(){
    char n1[N], n2[N];
    while(~scanf("%s%s", n1, n2)){
        string num1(n1), num2(n2);
        string ans = Mul(num1, num2);
        cout<

正解代码（使用FFT傅里叶变换，代码待补充，先给个链接：GO）