• 日期: 2026-02-28
  • 标签: 常规

首先，需要定义一个结构体来表示大整数：

#define MAX_LEN 1000  // 大整数最大长度

struct BigInt {
    int len;  // 大整数的长度
    char digits[MAX_LEN];  // 大整数的每一位，存放在字符数组中
};

接下来，实现大整数的加减乘除运算。

加法

大整数的加法可以看成是两个二进制数的加法，只需要从右往左逐位相加，记得处理进位即可。

// 大整数加法
struct BigInt add(struct BigInt a, struct BigInt b) {
    struct BigInt c = {0};
    int carry = 0;  // 进位
    int i, j;
    for (i = a.len - 1, j = b.len - 1; i >= 0 || j >= 0; i--, j--) {
        int x = i >= 0 ? a.digits[i] - '0' : 0;
        int y = j >= 0 ? b.digits[j] - '0' : 0;
        int sum = x + y + carry;
        c.digits[c.len++] = sum % 2 + '0';
        carry = sum / 2;
    }
    if (carry) {
        c.digits[c.len++] = carry + '0';
    }
    reverse(c.digits, c.digits + c.len);
    return c;
}

减法

大整数的减法可以看成是两个二进制数的减法，也是从右往左逐位相减，记得处理借位。

// 大整数减法
struct BigInt sub(struct BigInt a, struct BigInt b) {
    struct BigInt c = {0};
    int borrow = 0;  // 借位
    int i, j;
    for (i = a.len - 1, j = b.len - 1; i >= 0 || j >= 0; i--, j--) {
        int x = i >= 0 ? a.digits[i] - '0' : 0;
        int y = j >= 0 ? b.digits[j] - '0' : 0;
        int diff = x - y - borrow;
        if (diff < 0) {
            diff += 2;
            borrow = 1;
        } else {
            borrow = 0;
        }
        c.digits[c.len++] = diff + '0';
    }
    while (c.len > 1 && c.digits[c.len - 1] == '0') {
        c.len--;
    }
    reverse(c.digits, c.digits + c.len);
    return c;
}

乘法

大整数的乘法需要先将乘数和被乘数的每一位相乘，然后再将它们的乘积相加。这个过程中需要注意进位。

// 大整数乘法
struct BigInt mul(struct BigInt a, struct BigInt b) {
    struct BigInt c = {0};
    int i, j;
    for (i = a.len - 1; i >= 0; i--) {
        int carry = 0;  // 进位
        struct BigInt t = {0};
        for (j = b.len - 1; j >= 0; j--) {
            int x = a.digits[i] - '0';
            int y = b.digits[j] - '0';
            int product = x * y + carry;
            t.digits[t.len++] = product % 2 + '0';
            carry = product / 2;
        }
        if (carry) {
            t.digits[t.len++] = carry + '0';
        }
        reverse(t.digits, t.digits + t.len);
        for (j = 0; j < a.len - i - 1; j++) {
            t.digits[t.len++] = '0';
        }
        c = add(c, t);
    }
    return c;
}

除法

大整数的除法可以用二分查找来实现。首先将被除数左移，直到它的长度不小于除数的长度。然后每次将被除数减去除数的左移，直到被除数小于除数。这个过程中需要注意进位。

// 大整数除法
struct BigInt div(struct BigInt a, struct BigInt b) {
    struct BigInt c = {0};
    struct BigInt t = {0};
    int i, j;
    for (i = 0; i < a.len; i++) {
        t.digits[t.len++] = a.digits[i];
        if (compare(t, b) < 0) {
            c.digits[c.len++] = '0';
            continue;
        }
        int l = 1, r = 10;
        while (l < r) {
            int m = (l + r) >> 1;
            struct BigInt tmp = mul(b, {0, {(char)(m + '0')}});
            if (compare(tmp, t) <= 0) {
                l = m + 1;
            } else {
                r = m;
            }
        }
        c.digits[c.len++] = (char)(l - 1 + '0');
        struct BigInt tmp = mul(b, {0, {(char)(l - 1 + '0')}});
        t = sub(t, tmp);
    }
    while (c.len > 1 && c.digits[c.len - 1] == '0') {
        c.len--;
    }
    reverse(c.digits, c.digits + c.len);
    return c;
}

最后，附上完整的代码：

#include <stdio.h>
#include <string.h>

#define MAX_LEN 1000  // 大整数最大长度

struct BigInt {
    int len;  // 大整数的长度
    char digits[MAX_LEN];  // 大整数的每一位，存放在字符数组中
};

// 将字符数组反转
void reverse(char *begin, char *end) {
    while (begin < end) {
        char tmp = *begin;
        *begin = *end;
        *end = tmp;
        begin++;
        end--;
    }
}

// 比较两个大整数
int compare(struct BigInt a, struct BigInt b) {
    if (a.len != b.len) {
        return a.len - b.len;
    }
    for (int i = 0; i < a.len; i++) {
        if (a.digits[i] != b.digits[i]) {
            return a.digits[i] - b.digits[i];
        }
    }
    return 0;
}

// 大整数加法
struct BigInt add(struct BigInt a, struct BigInt b) {
    struct BigInt c = {0};
    int carry = 0;  // 进位
    int i, j;
    for (i = a.len - 1, j = b.len - 1; i >= 0 || j >= 0; i--, j--) {
        int x = i >= 0 ? a.digits[i] - '0' : 0;
        int y = j >= 0 ? b.digits[j] - '0' : 0;
        int sum = x + y + carry;
        c.digits[c.len++] = sum % 2 + '0';
        carry = sum / 2;
    }
    if (carry) {
        c.digits[c.len++] = carry + '0';
    }
    reverse(c.digits, c.digits + c.len);
    return c;
}

// 大整数减法
struct BigInt sub(struct BigInt a, struct BigInt b) {
    struct BigInt c = {0};
    int borrow = 0;  // 借位
    int i, j;
    for (i = a.len - 1, j = b.len - 1; i >= 0 || j >= 0; i--, j--) {
        int x = i >= 0 ? a.digits[i] - '0' : 0;
        int y = j >= 0 ? b.digits[j] - '0' : 0;
        int diff = x - y - borrow;
        if (diff < 0) {
            diff += 2;
            borrow = 1;
        } else {
            borrow = 0;
        }
        c.digits[c.len++] = diff + '0';
    }
    while (c.len > 1 && c.digits[c.len - 1] == '0') {
        c.len--;
    }
    reverse(c.digits, c.digits + c.len);
    return c;
}

// 大整数乘法
struct BigInt mul(struct BigInt a, struct BigInt b) {
    struct BigInt c = {0};
    int i, j;
    for (i = a.len - 1; i >= 0; i--) {
        int carry = 0;  // 进位
        struct BigInt t = {0};
        for (j = b.len - 1; j >= 0; j--) {
            int x = a.digits[i] - '0';
            int y = b.digits[j] - '0';
            int product = x * y + carry;
            t.digits[t.len++] = product % 2 + '0';
            carry = product / 2;
        }
        if (carry) {
            t.digits[t.len++] = carry + '0';
        }
        reverse(t.digits, t.digits + t.len);
        for (j = 0; j < a.len - i - 1; j++) {
            t.digits[t.len++] = '0';
        }
        c = add(c, t);
    }
    return c;
}

// 大整数除法
struct BigInt div(struct BigInt a, struct BigInt b) {
    struct BigInt c = {0};
    struct BigInt t = {0};
    int i, j;
    for (i = 0; i < a.len; i++) {
        t.digits[t.len++] = a.digits[i];
        if (compare(t, b) < 0) {
            c.digits[c.len++] = '0';
            continue;
        }
        int l = 1, r = 10;
        while (l < r) {
            int m = (l + r) >> 1;
            struct BigInt tmp = mul(b, {0, {(char)(m + '0')}});
            if (compare(tmp, t) <= 0) {
                l = m + 1;
            } else {
                r = m;
            }
        }
        c.digits[c.len++] = (char)(l - 1 + '0');
        struct BigInt tmp = mul(b, {0, {(char)(l - 1 + '0')}});
        t = sub(t, tmp);
    }
    while (c.len > 1 && c.digits[c.len - 1] == '0') {
        c.len--;
    }
    reverse(c.digits, c.digits + c.len);
    return c;
}

int main() {
    // 输入两个大整数
    char a_str[MAX_LEN], b_str[MAX_LEN];
    printf("请输入第一个大整数：");
    scanf("%s", a_str);
    printf("请输入第二个大整数：");
    scanf("%s", b_str);

    // 将字符串转换为大整数
    struct BigInt a = {0};
    for (int i = 0; a_str[i] != '\0'; i++) {
        a.digits[a.len++] = a_str[i];
    }
    struct BigInt b = {0};
    for (int i = 0; b_str[i] != '\0'; i++) {
        b.digits[b.len++] = b_str[i];
    }

    // 计算加减乘除结果
    struct BigInt sum = add(a, b);
    struct BigInt diff = sub(a, b);
    struct BigInt product = mul(a, b);
    struct BigInt quotient = div(a, b);

    // 输出结果
    printf("a + b = ");
    for (int i = 0; i < sum.len; i++) {
        printf("%c", sum.digits[i]);
    }
    printf("\n");

    printf("a - b = ");
    for (int i = 0; i < diff.len; i++) {
        printf("%c", diff.digits[i]);
    }
    printf("\n");

    printf("a * b = ");
    for (int i = 0; i < product.len; i++) {
        printf("%c", product.digits[i]);
    }
    printf("\n");

    printf("a / b = ");
    for (int i = 0; i < quotient.len; i++) {
        printf("%c", quotient.digits[i]);
    }
    printf("\n");

    return 0;
}
C语言实现大整数加减乘除运算：二进制字符数组存储

原文地址: https://www.cveoy.top/t/topic/nJvF 著作权归作者所有。请勿转载和采集!

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