高精度入门（减法、加法、乘法）之 CODE[VS] 3115、3116、3117

3115 高精度练习之减法：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <string>
#include <algorithm>
#include <cstring>
#include <set>
#include <utility>
#include <locale>
#include <ctime>
using namespace std;
//using int64 = long long;
const int INF = 0x3f3f3f3f;
const int MaxN = 110;

string A, B;
string ans;

void Repair()
{
    // 使两个字符串具有相同长度
    if (A.size() < B.size())
    {
        for (int i = A.size(); i < B.size(); ++i)
        {
            A = '0' + A;
        }
    }else
    {
        for (int i = B.size(); i < A.size(); ++i)
        {
            B = '0' + B;
        }
    }
}

void Solve()
{

    bool isNegative = true;
    // 倘若A<B，输出负号，并交换A和B的值
    for (int i = 0; i < A.size(); ++i)
    {
        if (A[i] - '0' == B[i] - '0') continue;
        if (A[i]-'0' > B[i]-'0')
        {
            isNegative = false;
        }
        break;
    }
    if (isNegative) {
        cout << "-";
        swap(A, B);
    }
    // 大数减法
    for (int i = A.size() - 1; i >= 0; --i)
    {
        if (A[i] < B[i])
        {
            // 借位，注意所借之位为0的情况，借位之后，这些位变为9
            int tmp = i - 1;
            while ((A[tmp]-'0')-1 < 0) --tmp;
            A[tmp] = '0' + ((A[tmp] - '0') - 1);
            ++tmp;
            while (tmp < i)
            {
                A[tmp] = '9';
                ++tmp;
            }
            ans = static_cast<char>('0' + ((A[i] - '0' + 10) - (B[i] - '0'))) + ans;
        }else
        {
            // 无需借位，直接减
            ans = static_cast<char>('0' + (A[i] - B[i])) + ans;
        }
    } 
    bool Judge = false;
    for (int i = 0; i < ans.size(); ++i)
    {
        // 忽略前导零
        if (ans[i] != '0' || Judge)
        {
            cout << ans[i];
            Judge = true;
        }
    }
    cout << endl;
}

int main()
{

    while (cin >> A >> B)
    {
        ans.clear();
        Repair();
        Solve();
    }

#ifdef HOME
    cerr << "Time elapsed: " << clock() / CLOCKS_PER_SEC << " ms" << endl;
#endif
    return 0;
}

3116 高精度练习之加法：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <string>
#include <algorithm>
#include <cstring>
#include <set>
#include <utility>
#include <locale>
#include <ctime>
using namespace std;
//using int64 = long long;
const int INF = 0x3f3f3f3f;
const int MaxN = 110;

string A, B;
string ans;

void Repair()
{
    if (A.size() < B.size())
    {
        for (int i = A.size(); i < B.size(); ++i)
        {
            A = '0' + A;
        }
    }else
    {
        for (int i = B.size(); i < A.size(); ++i)
        {
            B = '0' + B;
        }
    }
}

void Solve()
{
    int tmp = 0;
    for (int i = A.size() - 1; i >= 0; --i)
    {
        if ((A[i]-'0') + (B[i] - '0') + tmp > 9)
        {
            ans = static_cast<char>('0' + ((A[i] - '0') + (B[i] - '0') + tmp) % 10) + ans;
            tmp = 1;
        }else
        {
            ans = static_cast<char>('0' + ((A[i] - '0') + (B[i] - '0') + tmp)) + ans;
            tmp = 0;
        }
    }
    if (tmp) ans = '1' + ans;
    cout << ans << endl;;
}

int main()
{

    while (cin >> A >> B)
    {
        ans.clear();
        Repair();
        Solve();
    }

#ifdef HOME
    cerr << "Time elapsed: " << clock() / CLOCKS_PER_SEC << " ms" << endl;
#endif
    return 0;
}

3117 高精度练习之乘法：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <string>
#include <algorithm>
#include <cstring>
#include <set>
#include <utility>
#include <locale>
#include <ctime>
using namespace std;
//using int64 = long long;
const int INF = 0x3f3f3f3f;
const int MaxN = 110;

string _A, _B;

void Repair(string& A, string& B)
{
    if (A.size() < B.size())
    {
        for (int i = A.size(); i < B.size(); ++i)
        {
            A = '0' + A;
        }
    }else
    {
        for (int i = B.size(); i < A.size(); ++i)
        {
            B = '0' + B;
        }
    }
}

string Add(string A, string B)
{
    Repair(A, B);
    string ans;
    int tmp = 0;
    for (int i = A.size() - 1; i >= 0; --i)
    {
        if ((A[i] - '0') + (B[i] - '0') + tmp > 9)
        {
            ans = static_cast<char>('0' + ((A[i] - '0') + (B[i] - '0') + tmp) % 10) + ans;
            tmp = 1;
        }
        else
        {
            ans = static_cast<char>('0' + ((A[i] - '0') + (B[i] - '0') + tmp)) + ans;
            tmp = 0;
        }
    }
    if (tmp) ans = '1' + ans;
    return ans;
}

string Mul(string A, char B)
{
    string ans;
    int tmp = 0;
    int result;
    for (int i = A.size() - 1; i >= 0; --i)
    {
        result = (A[i] - '0')*(B - '0') + tmp;
        tmp = result / 10;
        ans = static_cast<char>('0' + result % 10) + ans;
    }
    if (tmp) ans = static_cast<char>('0' + tmp) + ans;
    return ans;
}

void Solve(string A, string B)
{
    string ans = "0";
    string tmp;
    int num = 0; // 统计加的0的个数
    for (int i = B.size() - 1; i >= 0; --i)
    {
        tmp = Mul(A, B[i]);
        for (int j = 0; j < num; ++j) tmp += '0';
        ans = Add(ans, tmp);
        ++num;
    }
    cout << ans << endl;;
}

int main()
{

    while (cin >> _A >> _B)
    {
        Solve(_A, _B);
    }

#ifdef HOME
    cerr << "Time elapsed: " << clock() / CLOCKS_PER_SEC << " ms" << endl;
#endif
    return 0;
}