高精度减法

重点
1. t 处理借位，开始是0
2. c的高位需要判断是否为0，全部pop，只剩1位时，保留
3.

#include <iostream>
#include <cstring>
#include <algorithm>
#include <vector>

using namespace std;
const int N = 1000010;

int cmp(vector<int> &a, vector<int> &b) {
    if (a.size() != b.size()) return a.size() > b.size();
    for (int i = a.size() - 1; i >= 0; i--) {
        if (a[i] != b[i]) return a[i] > b[i];
    }
    return 1;
}

vector<int> sub(vector<int> &a, vector<int> &b) {
    vector<int> c;
    int t = 0;
    for (int i = 0; i < a.size(); i++) {
        t = a[i] - t;
        if (i < b.size()) t -= b[i];
        c.push_back((t + 10) % 10);
        if (t < 0) t = 1;
        else t = 0;
    }
    while (c.size() > 1 && c.back() == 0) c.pop_back();
    return c;
}

int main() {
    string s1, s2;
    cin >> s1 >> s2;
    vector<int> a, b;
    for (int i = s1.size() - 1; i >= 0; i--) a.push_back(s1[i] - '0');
    for (int i = s2.size() - 1; i >= 0; i--) b.push_back(s2[i] - '0');

    if (cmp(a, b)) {
        auto c = sub(a, b);
        for (int i = c.size() - 1; i >= 0;  i--) cout << c[i];
    } else {
        auto c = sub(b, a);
        cout << '-';
        for (int i =  c.size() - 1; i >= 0; i--) cout << c[i];
    }

    return 0;
}