【高精度减法】Acwing792.高精度减法

Acwing792.高精度减法

题解

#include <iostream>
#include <vector>
using namespace std;

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;
}

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


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

压九位写法

#include <iostream>
#include <vector>
#include <cstdio>

using namespace std;

const int M = 1000000000;

bool cmp(vector<int> &A, vector<int> &B)
{
    if(A.size() > B.size()) return true;
    else if(A.size() < B.size()) return false;
    
    for(int i = A.size() - 1; i >= 0; --i)
        if(A[i] != B[i])
            return A[i] > B[i];
    return true;
}

vector<int> sub(vector<int> &A, vector<int> &B)
{
    if(!cmp(A, B)) {
        cout << "-";
        return sub(B, A);
    }
    vector<int> C;
    int s = 0, r = 0;
    for(int i = 0; i < A.size(); ++i)
    {
        s = A[i] + r;
        r = 0;
        if(i < B.size())
            s -= B[i];
        if(s < 0)
            s += M, r = -1;
        C.push_back(s % M);
    }
    while(C.size() && C.back() == 0) C.pop_back();
    return C;
}

int main()
{
    string a, b;
    vector<int> A, B;
    cin >> a >> b;
    
    for(int i = a.size() - 1, j = 0, t = 1, x = 0; i >= 0; --i)
    {
        x += (a[i] - '0') * t;
        t *= 10, j++;
        if(j == 9 || i == 0)
        {
            A.push_back(x);
            x = 0, t = 1, j = 0;
        }
    }
    for(int i = b.size() - 1, j = 0, t = 1, x = 0; i >= 0; --i)
    {
        x += (b[i] - '0') * t;
        t *= 10, j++;
        if(j == 9 || i == 0)
        {
            B.push_back(x);
            x = 0, t = 1, j = 0;
        }
    }
    
    auto C = sub(A, B);
    printf("%d", C[C.size() - 1]);
    for(int i = C.size() - 2; i >= 0; --i)
        printf("%09d", C[i]);
    
    return 0;
}