高精度

前置知识

数组，模拟

目标

高精加高精，高精减高精，高精乘低精，高精乘高精，高精除低精

高精加高精

//一维数组版本

#include <bits/stdc++.h>

using namespace std;

const int N = 220;

int A[N], B[N], C[N], lena, lenb, lenc;
string a, b;

void add(int A[], int B[], int C[]){
    int t = 0;
    for (int i = 0; i < lena || i < lenb; i++){
        if (i < lena) t += A[i];
        if (i < lenb) t += B[i];
        C[lenc++] = t % 10;
        t /= 10;
    }
    if (t) C[lenc++] = 1;

    while (lenc > 0 && C[lenc] == 0) lenc--;
    for (int i = lenc; i >= 0; i--) cout << C[i];
    puts("");
}

int main(){
    cin >> a >> b;

    for (int i = a.size() - 1; i >= 0; i--) A[lena++] = a[i] - '0';
    for (int i = b.size() - 1; i >= 0; i--) B[lenb++] = b[i] - '0';

    add(A, B, C);

    return 0;
}

// C = A + B, A >= 0, B >= 0
vector<int> add(vector<int> &A, vector<int> &B)
{
    if (A.size() < B.size()) return add(B, A);

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

    if (t) C.push_back(t);
    return C;
}

高精减高精

// 一维数组版本
#include <bits/stdc++.h>

using namespace std;

const int N = 220;

int A[N], B[N], C[N], lena, lenb, lenc;
string a, b;

void sub(int A[], int B[], int C[]){
    int t = 0;
    for (int i = 0; i < lena; i++){
        t = A[i] - t;
        if (i < lenb) t -= B[i];

        C[lenc++] = (t + 10) % 10;

        if (t < 0) t = 1;
        else t = 0;
    }

    while (lenc > 0 && C[lenc] == 0) lenc--;
    for (int i = lenc; i >= 0; i--) cout << C[i];
    puts("");
}

int main(){
    cin >> a >> b;

    for (int i = a.size() - 1; i >= 0; i--) A[lena++] = a[i] - '0';
    for (int i = b.size() - 1; i >= 0; i--) B[lenb++] = b[i] - '0';

    sub(A, B, C);

    return 0;
}

// C = A - B, 满足A >= B, A >= 0, B >= 0
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] - 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;
}

高精乘低精

//一维数组版本
#include <bits/stdc++.h>

using namespace std;

const int N = 220;

int A[N], C[N], lena, lenc;
string a;
int b;

void mul(int A[], int b, int C[]){
    int t = 0;
    for (int i = 0; i < lena || t; i++) {
        if (i < lena) t += A[i] * b;
        C[lenc++] = t % 10;
        t /= 10;
    }

    while (lenc > 0 && C[lenc] == 0) lenc--;

    for (int i = lenc; i >= 0; i--) cout << C[i];
    puts("");
}

int main(){
    cin >> a >> b;

    for (int i = a.size() - 1; i >= 0; i--) A[lena++] = a[i] - '0';

    mul(A, b, C);

    return 0;
}   

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

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

    while (C.size() > 1 && C.back() == 0) C.pop_back();

    return C;
}

高精乘高精

//一维数组版本
#include <bits/stdc++.h>

using namespace std;

const int N = 550;

int A[N], B[N], C[N];
int lena, lenb, lenc;
string a, b;

void mul(int A[], int B[], int C[]){
    for (int i = 0; i < lena; i++)
        for (int j = 0; j < lenb; j++)
            C[i + j] += A[i] * B[j];

    lenc = lena + lenb;
    for (int i = 0; i < lenc; i++){
        C[i + 1] += C[i] / 10;
        C[i] %= 10;
    }

    while (lenc > 0 && C[lenc] == 0) lenc--;
    for (int i = lenc; i >= 0; i--) cout << C[i];
    puts("");
}

int main(){
    cin >> a >> b;
    for (int i = a.size() - 1; i >= 0; i--) A[lena++] = a[i] - '0';
    for (int i = b.size() - 1; i >= 0; i--) B[lenb++] = b[i] - '0';

    mul(A, B, C);

    return 0;
}

vector<int> mul(vector<int> A, vector<int> B)
{
    vector<int> C(A.size() + B.size());

    for (int i = 0; i < A.size(); i ++ )
        for (int j = 0; j < B.size(); j ++ )
            C[i + j] += A[i] * B[j];

    for (int i = 0, t = 0; i < C.size() || t; i ++ )
    {
        t += C[i];
        if (i >= C.size()) C.push_back(t % 10);
        else C[i] = t % 10;
        t /= 10;
    }

    while (C.size() > 1 && !C.back()) C.pop_back();

    return C;
}

高精除低精

//一维数组版本
#include <bits/stdc++.h>

using namespace std;

const int N = 220;

int A[N], C[N], lena, lenc;

void div(int A[], int b, int C[]){
    int t = 0;
    for (int i = 0; i < lena; i++){
        t = t * 10 + A[i];
        C[i] = t / b;
        t %= b;
    }

    while (lenc < lena && C[lenc] == 0) lenc++;

    if (lenc == lena) lenc--;
    for (int i = lenc; i < lena; i++) cout << C[i];
    puts("");
    cout << t << '\n';
}

int main(){
    string a;
    int b = 13;
    cin >> a;

    for (int i = 0; i < a.size(); i++) A[lena++] = a[i] - '0';

    div(A, b, C);

    return 0;
}

// vector版本
// A / b = C ... r, A >= 0, b > 0
vector<int> div(vector<int> &A, int b, int &r)
{
    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;
}

题单

1168：大整数加法

1169：大整数减法

1170：计算2的N次方

1174：大整数乘法

1175：除以13

总结

高精度属于模拟算法，重点考察代码能力

在比赛中一般会开long long得部分分

但对于专业选手，在代码的某个环节，转换成高精度，写起来不费劲，是一个基本功