大整数运算练习

#include<stdio.h>
#include<string.h>
struct bign {
    int d[1000];
    int len;
    bign() {   //构造函数（函数名和结构体相同， 无返回值）
        memset(d, 0, sizeof(d)); //头文件<string.h>
        len = 0;
    }
};

bign change(char str[]) {   //字符串倒着赋给d[]数组
    bign a;
    a.len = strlen(str);
    for (int i = 0; i < a.len; i++) {
        a.d[i] = str[a.len - i - 1] - ‘0‘;  //逆着赋值
    }
    return a;
}

int compare(bign a, bign b) {   //比较两个bign变量的大小
    if (a.len > b.len) return 1;
    else if (a.len < b.len) return -1;
    else {
        for (int i = a.len - 1; i >= 0; i--) {
            if (a.d[i] > b.d[i]) return 1;
            else if (a.d[i] < b.d[i]) return -1;
        }
        return 0;
    }
}
bign add(bign a, bign b) {  //高精度加法
    bign c;
    int carry = 0;  //进位
    for (int i = 0; i < a.len || i < b.len; i++) {
        int temp = a.d[i] + b.d[i] + carry;
        c.d[c.len++] = temp % 10;
        carry = temp / 10;
    }
    if (carry != 0) {
        c.d[c.len++] = carry;
    }
    return c;
}
bign sub(bign a, bign b) {   //高精度减法（使用前需要比较两个数的大小， 如果被减数小于减数， 需要交换两个变量，然后输出负号， 再使用sub函数）
    bign c;
    for (int i = 0; i < a.len || i < b.len; i++) {
        if (a.d[i] < b.d[i]) {
            a.d[i + 1]--;
            a.d[i] += 10;
        }
        c.d[c.len++] = a.d[i] - b.d[i];
    }
    while (c.len - 1 >= 1 && c.d[c.len - 1] == 0) {
        c.len--;
    }
    return c;
}

bign multi(bign a, int b) { //高精度与低精度的乘法
    bign c;
    int carry = 0;
    for (int i = 0; i < a.len; i++) {
        int temp = a.d[i] * b + carry;
        c.d[a.len++] = temp % 10;
        carry = temp / 10;
    }
    while (carry != 0) {
        c.d[c.len++] = carry % 10;
        carry /= 10;
    }
    return c;
}
bign divide(bign a, int b, int &r) {
    bign c;
    c.len = a.len;  //被除数的每一位和商的每一位是一一对应的。因此先令长度相等
    for (int i = a.len - 1; i >= 0; i--) {
        r = r * 10 + a.d[i];
        if (r < b) c.d[i] = 0;
        else {
            c.d[i] = r / b;  //商
            r = r % b;  //获得的余数
        }
    }
    while (c.len - 1 >= 1 && c.d[c.len - 1] == 0) {
        c.len--;
    }
    return c;
}
bign divide(bign a, int b, int &r) {   //高精度除法与低精度除法的运算  // r为余数
    bign c;
    c.len = a.len;
    for (int i = a.len - 1; i >= 0; i--) {
        r = r * 10 + a.d[i];
        if (r < b) c.d[i] = 0;
        else {
            c.d[i] = r / b;
            r = r % b;
        }
    }
    while (c.len - 1 >= 1 && c.d[c.len - 1] == 0) {
        c.len--;
    }
    return c;

}

int main() {
    char str1[1000], str2[1000];
    scanf("%s%s", str1, str2);
    bign a = change(str1);
    bign b = change(str2);
    bign c = add(a, b);
    for (int i = c.len - 1; i >= 0; --i) {
        printf("%d", c.d[i]);
    }
    return 0;
}

原文地址：https://www.cnblogs.com/Yaxadu/p/9189309.html