题解 | #Powerful Calculator#

本题用例无须考虑负数，若有负数仍可以用这些基运算实现

#include <iostream>
#include "cmath"
using namespace std;
struct bigInteger {
    int nums[1000];
    int length;
    bigInteger();
    bigInteger(string str);
    bool operator<( bigInteger B);
    bigInteger operator+(bigInteger B);
    bigInteger operator-(bigInteger B);
    bigInteger operator*(bigInteger B);
    void print();
};
bigInteger::bigInteger() {
    for (int i = 0; i < 1000; i++) {
        nums[i] = 0;
    }
    length = 0;
}
bigInteger::bigInteger(string str) {
    for (int i = 0; i < 1000; i++) {
        nums[i] = 0;
    }
    length = str.size();
    for (int i = 0; i < length; i++) {
        nums[i] = str[length - 1 - i] - '0';
    }
}
bool bigInteger::operator<(bigInteger B) {
    if (length < B.length) return true;
    else if (length > B.length) return false;
    else {
        for (int i = length - 1; i >= 0; i--) {
            if (nums[i] < B.nums[i]) return true;
            else if (nums[i] > B.nums[i]) return false;

        }
        return false;//全部数字相等
    }
}
bigInteger bigInteger::operator+(bigInteger B) {
    bigInteger ans;
    int carry = 0;
    int bound = max(length, B.length);
    for (int i = 0; i < bound ; i++) {
        int temp = nums[i] + B.nums[i] + carry;
        ans.nums[i] = temp % 10;
        ans.length++;
        carry = temp / 10;
    }
    if (carry != 0) ans.nums[ans.length++] = 1;

    return ans;
}
bigInteger bigInteger::operator-(bigInteger B) {
    bigInteger ans;
    int carry = 0;
    for (int i = 0; i < length; i++) {
        int temp = nums[i] - carry - B.nums[i];
        if (temp < 0) {
            temp += 10;
            carry = 1;
        } else {
            carry = 0;
        }
        ans.nums[ans.length++] = temp;
    }
    while (ans.nums[ans.length - 1] == 0 && ans.length >= 1) {
        ans.length--;
    }
    return ans;
}
bigInteger bigInteger::operator*(bigInteger B) {
    bigInteger ans;
    ans.length=length+B.length;
    for (int i = 0; i < B.length; i++) {
        for (int j = 0; j < length; j++) {
            ans.nums[i + j] += B.nums[i] * nums[j];
        }
    }
    int carry = 0;
    for (int i = 0; i < length + B.length; i++) {
        carry = ans.nums[i] / 10;
        ans.nums[i+1]+=carry;
        ans.nums[i] %= 10;
    }
    while (ans.nums[ans.length - 1] == 0 && ans.length >= 1) {
        ans.length--;
    }
    return ans;
}
void bigInteger::print() {
    for (int i = length - 1; i >= 0; i--) {
        cout << nums[i];
    }
    cout << endl;
}
int main() {
    string a, b;
    while (cin >> a >> b) { // 注意 while 处理多个 case
        // cout << bigIa + b << endl;
(bigInteger(a)+bigInteger(b)).print();
bigInteger A=bigInteger(a);
bigInteger B=bigInteger(b);
if(B<A)
(bigInteger(a)-bigInteger(b)).print();
else{cout<<'-';
(B-A).print();
}
(bigInteger(a)*bigInteger(b)).print();

    }
}
// 64 位输出请用 printf("%lld")