【基础算法】高精度专题

高精度模板

加法

vector<int> add(vector<int>& A, vector<int>& B) // 高精度加法
{
    vector<int> C;
    int t = 0;
    for (int i = 0; i < A.size() || i < B.size(); i ++ )
    {
        if (i < A.size()) t += A[i];
        if (i < B.size()) t += B[i];
        C.push_back(t % 10);
        t /= 10;
    }
    if (t) C.push_back(1);
    return C;
}

减法

bool cmp(vector<int>& A, vector<int>& B) // 判断A >= 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) // 高精度减法
{
    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;
}

乘法

大数乘小数

// C = A * b
vector<int> mul(vector<int> &A,int b)
{
    vector<int> C;

    int t = 0;
    for(int i = 0; i < A.size(); i++)
    {
        if(i < A.size()) t += A[i] * b;
        C.push_back(t % 10);
        t /= 10;
    }
    if(t) C.push_back(t);

    while(C.size() > 1 && C.back() == 0) C.pop_back();//去掉前导0

    return C;
}

大数乘大数

vector<int> mul(vector<int>& A, vector<int>& B) // 高精度乘法
{
    vector<int> C(A.size() + B.size(), 0);
    for (int i = 0; i < A.size(); i ++ )
        for (int j = 0; j < B.size(); j ++ )
            C[i + j] += A[i] * B[j];

    int t = 0;
    for (int i = 0; i < C.size(); i ++ )
    {
        t += C[i];
        C[i] = t % 10;
        t /= 10;
    }
    while (C.size() > 1 && C.back() == 0) C.pop_back();
    return C;
}

除法

// A / b 商是C 余数是r
vector<int> div(vector<int> &A, int b, int &r)// 这里r传的是引用
{
    vector<int> C;
    r = 0;
    for (int i = A.size() - 1; i >= 0; i -- )
    {
        r = r * 10 + A[i];
        C.push_back(r / b);
        r %= b;
    }
    reverse(C.begin(),C.end());
    while (C.size() > 1 && C.back() == 0) C.pop_back();
    return C;
}

3389. N 的阶乘

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

using namespace std;

const int N = 1010;

vector<int> F[N];

vector<int> mul(vector<int> &A, int b)
{
    vector<int> C;
    for(int i = 0, t = 0; i < A.size() || t; i ++ )
    {
        if(i < A.size()) t += A[i] * b;
        C.push_back(t % 10);
        t /= 10;
    }
    return C;
}

int main()
{
    int n;
    F[0] = {1};
    for(int i = 1; i <= 1000; i ++ ) F[i] = mul(F[i - 1], i);

    while(cin >> n)
    {
        for(int i = F[n].size() - 1; i >= 0; i -- )
        {
            cout << F[n][i];
        }
        cout << endl;
    }

    return 0;
}

作者:NFYD
链接:https://www.acwing.com/activity/content/code/content/3006568/
来源:AcWing
著作权归作者所有。商业转载请联系作者获得授权,非商业转载请注明出处。

3448. 基本算术

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

using namespace std;

int add(vector<int> &A, vector<int> &B)
{
    int res = 0;
    for(int i = 0, t = 0; i < A.size() || i < B.size() || t; i ++ )
    {
        if(i < A.size()) t += A[i];
        if(i < B.size()) t += B[i];
        t /= 10;
        res += t;
    }
    return res;
}

int main()
{
    string a, b;
    while(cin >> a >> b, a != "0" || b != "0")
    {
        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(add(A, B) == 0) printf("No carry operation.\n");
        else if(add(A, B) == 1) printf("%d carry operation.\n", add(A, B));
        else printf("%d carry operations.\n", add(A, B));
    }
    return 0;
}

作者:NFYD
链接:https://www.acwing.com/activity/content/code/content/3009963/
来源:AcWing
著作权归作者所有。商业转载请联系作者获得授权,非商业转载请注明出处。

3453. 整数查询

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

using namespace std;

vector<int> add(vector<int> &A, vector<int> &B)
{
    vector<int> C;
    for(int i = 0, t = 0; i < A.size() || i < B.size() || t; i ++ )
    {
        if(i < A.size()) t += A[i];
        if(i < B.size()) t += B[i];
        C.push_back(t % 10);
        t /= 10;
    }
    return C;
}

int main()
{
    vector<int> A{0};
    string b;
    while(cin >> b, b != "0")
    {
        vector<int> B;
        for(int i = b.size() - 1; i >= 0; i -- ) B.push_back(b[i] - '0');
        A = add(A, B);
    }
    while(A.size() > 1 && A.back() == '0') A.pop_back(); //去掉前导零
    for(int i = A.size() - 1; i >= 0; i -- ) cout << A[i];
    cout << endl;
    return 0;
}

作者:NFYD
链接:https://www.acwing.com/activity/content/code/content/3014579/
来源:AcWing
著作权归作者所有。商业转载请联系作者获得授权,非商业转载请注明出处。

3482. 大数运算

#include <iostream>
#include <vector>

using namespace std;

vector<int> add(vector<int>& A, vector<int>& B) // 高精度加法
{
    vector<int> C;
    int t = 0;
    for (int i = 0; i < A.size() || i < B.size(); i ++ )
    {
        if (i < A.size()) t += A[i];
        if (i < B.size()) t += B[i];
        C.push_back(t % 10);
        t /= 10;
    }
    if (t) C.push_back(1);
    return C;
}

bool cmp(vector<int>& A, vector<int>& B) // 判断A >= 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) // 高精度减法
{
    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;
}

vector<int> mul(vector<int>& A, vector<int>& B) // 高精度乘法
{
    vector<int> C(A.size() + B.size(), 0);
    for (int i = 0; i < A.size(); i ++ )
        for (int j = 0; j < B.size(); j ++ )
            C[i + j] += A[i] * B[j];
    
    int t = 0;
    for (int i = 0; i < C.size(); i ++ )
    {
        t += C[i];
        C[i] = t % 10;
        t /= 10;
    }
    while (C.size() > 1 && C.back() == 0) C.pop_back();
    return C;
}

void print(vector<int>& C) // 因为要多次输出,所以封装成函数
{
    for (int i = C.size() - 1; i >= 0; i -- ) printf("%d", C[i]);
    puts("");
    C.clear();
}

int main()
{
    string a, b;
    cin >> a >> b;
    vector<int> A, B;
    
    for (int i = a.size() - 1; i >= 0; i -- ) 
        if (a[i] != '-')
            A.push_back(a[i] - '0');
            
    for (int i = b.size() - 1; i >= 0; i -- ) 
        if (b[i] != '-')
            B.push_back(b[i] - '0');
    
    vector<int> C;
    
    if (a[0] != '-' && b[0] != '-') // a + b, a - b, a * b
    {
        C = add(A, B);
        print(C);
        
        if (cmp(A, B)) C = sub(A, B);
        else 
        {
            C = sub(B, A);
            printf("-");
        }
        print(C);
        
        C = mul(A, B);
        print(C);
    }
    else if (a[0] == '-' && b[0] == '-') // (-a) + (-b), (-a) - (-b), (-a) * (-b)
    {
        C = add(A, B);
        printf("-");
        print(C);
        
        if (cmp(A, B))
        {
            printf("-");
            C = sub(A, B);
            print(C);
        }
        else
        {
            C = sub(B, A);
            print(C);
        }
        
        C = mul(A, B);
        print(C);
    }
    else if (a[0] == '-' && b[0] != '-') // (-a) + b, (-a) - b, (-a) * b
    {
        if (cmp(A, B)) 
        {
            C = sub(A, B);
            printf("-");
        }
        else C = sub(B, A);
        print(C);
        
        printf("-");
        C = add(A, B);
        print(C);
        
        printf("-");
        C = mul(A, B);
        print(C);
    }
    else if (a[0] != '0' && b[0] == '-') // a + (-b), a - (-b), a * (-b)
    {
        if (cmp(A, B)) C = sub(A, B);
        else
        {
            C = sub(B, A);
            printf("-");
        }
        print(C);
        
        C = add(A, B);
        print(C);
        
        printf("-");
        C = mul(A, B);
        print(C);
    }
    return 0;
}
posted @   Tshaxz  阅读(26)  评论(0编辑  收藏  举报
相关博文:
阅读排行:
· 在鹅厂做java开发是什么体验
· 百万级群聊的设计实践
· WPF到Web的无缝过渡:英雄联盟客户端的OpenSilver迁移实战
· 永远不要相信用户的输入:从 SQL 注入攻防看输入验证的重要性
· 全网最简单!3分钟用满血DeepSeek R1开发一款AI智能客服,零代码轻松接入微信、公众号、小程
Language: HTML
点击右上角即可分享
微信分享提示