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高精度减法模板

大数减法:借位处理与正负判断

高精度 减法 大数 借位 GESP5
#include <bits/stdc++.h>
using namespace std;

// 比较 A >= B
bool 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 true;
}

// 高精度减法:C = A - B(保证 A >= B)
vector<int> sub(vector<int> &A, vector<int> &B) {
    vector<int> C;
    int borrow = 0;
    for (int i = 0; i < A.size(); i++) {
        int t = A[i] - borrow;
        if (i < B.size()) t -= B[i];
        C.push_back((t + 10) % 10);
        borrow = t < 0 ? 1 : 0;
    }
    while (C.size() > 1 && C.back() == 0) C.pop_back();  // 去前导零
    return C;
}

int main() {
    string a, b;
    cin >> a >> b;
    vector<int> A, B;
    for (int i = a.size()-1; i >= 0; i--) A.push_back(a[i]-'0');
    for (int i = b.size()-1; i >= 0; i--) B.push_back(b[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];
    }
    cout << endl;
    return 0;
}

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