洛谷题单-高精度

加法

传送门

#include <bits/stdc++.h>
using namespace std;
// https://www.luogu.com.cn/problem/P1601
char n1[10000], n2[10000];
int num1[10000], num2[10000];
int ans[10000];
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);
    cin >> n1;
    cin >> n2;
    int len1 = strlen(n1);
    int len2 = strlen(n2);
    memset(num1, 0, sizeof(num1));
    memset(num2, 0, sizeof(num2));
    memset(ans, 0, sizeof(ans));
    for (int i = 1; i <= len1; i++)
    {
        num1[i] = n1[len1 - i] - '0';
        // cout << num1[i];
    }
    // cout << endl;
    for (int i = 1; i <= len2; i++)
    {
        num2[i] = n2[len2 - i] - '0';
        // cout << num2[i];
    }
    // cout << endl;
    int lenc = 1;
    int mod1 = 0;
    while (lenc <= len1 || lenc <= len2)
    {
        ans[lenc] = num1[lenc] + num2[lenc] + mod1;
        mod1 = ans[lenc] / 10;
        ans[lenc] %= 10;
        lenc++;
    }
    ans[lenc] = mod1;
    if (ans[lenc] == 0)
        lenc--; //防止上一位0
    for (int i = lenc; i > 0; i--)
    {
        cout << ans[i];
    }
    return 0;
}

减法

传送门

#include <bits/stdc++.h>
using namespace std;
// https://www.luogu.com.cn/problem/P2142
char n1[1000000], n2[1000000];
char remp[1000000];
int num1[1000000], num2[1000000];
int ans[1000000];
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);
    cin >> n1 >> n2;
    if (strlen(n1) < strlen(n2) || (strlen(n1) == strlen(n2) && strcmp(n1, n2) < 0))
    {
        //当被减数小于减数时，两者交换，且加上“-”
        strcpy(remp, n1);
        strcpy(n1, n2);
        strcpy(n2, remp);
        cout << "-";
    }
    int len1 = strlen(n1);
    int len2 = strlen(n2);
    for (int i = 1; i <= len1; i++)
    {
        num1[i] = n1[len1 - i] - '0';
    }
    for (int i = 1; i <= len2; i++)
    {
        num2[i] = n2[len2 - i] - '0';
    }

    int lenc = 1;
    while (lenc <= len1 || lenc <= len2)
    {
        if (num1[lenc] < num2[lenc])
        {
            num1[lenc] += 10;
            num1[lenc + 1]--;
        }
        ans[lenc] = num1[lenc] - num2[lenc];
        // cout << num1[lenc] << " " << num2[lenc] << " " << ans[lenc] << " " << lenc << endl;
        lenc++;
    }
    while (ans[lenc] == 0 && lenc > 1) // lenc大于1不是大于等于！
    {
        lenc--;
    }
    for (int i = lenc; i >= 1; i--)
    {
        cout << ans[i];
    }
}

乘法

传送门

#include <bits/stdc++.h>
using namespace std;
char n1[1000000], n2[1000000];
int num1[1000000], num2[1000000], ans[1000000];
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);
    cin >> n1;
    cin >> n2;
    int len1 = strlen(n1);
    int len2 = strlen(n2);
    for (int i = 1; i <= len1; i++)
    {
        num1[i] = n1[len1 - i] - '0';
    }
    for (int i = 1; i <= len2; i++)
    {
        num2[i] = n2[len2 - i] - '0';
    }
    for (int i = 1; i <= len1; i++)
    {
        int mod1 = 0; //用来进位
        for (int j = 1; j <= len2; j++)
        {
            ans[i + j - 1] += num1[i] * num2[j] + mod1;
            mod1 = ans[i + j - 1] / 10;
            ans[i + j - 1] %= 10;
        }
        ans[i + len2] = mod1;
    }
    int lenc = len1 + len2;
    while (lenc > 1 && ans[lenc] == 0)
        lenc--;
    for (int i = lenc; i > 0; i--)
        cout << ans[i];
    return 0;
}

除法

传送门

#include <bits/stdc++.h>
using namespace std;
char n1[1000000];
int num1[1000000], ans[1000000];
long long num2;
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);
    cin >> n1;
    cin >> num2;
    int len1 = strlen(n1);
    for (int i = 1; i <= len1; i++)
    {
        num1[i] = n1[i - 1] - '0';
    }
    long long mod1 = 0;
    for (int i = 1; i <= len1; i++)
    {
        ans[i] = (mod1 * 10 + num1[i]) / num2;
        // cout<<ans[i]<<endl;
        mod1 = (mod1 * 10 + num1[i]) % num2;
    }
    int lenc = 1;
    while (lenc < len1 && ans[lenc] == 0)
        lenc++;

    for (int i = lenc; i <= len1; i++)
        cout << ans[i];
    return 0;
}