用C++写一个高精度加减乘除

我可以给出一个示例：#include <stdio.h>

int main()
{
double a, b, c;
char op;

printf("请输入算式（例如：2 + 3）: ");
scanf("%lf %c %lf", &a, &op, &b);

switch(op)
{
case '+':
c = a + b;
break;

case '-':
c = a - b;
break;

case '*':
c = a * b;
break;

case '/':
c = a / b;
break;

default:
printf("无效的运算符\n");
break;
}

printf("%.2lf %c %.2lf = %.2lf\n", a, op, b, c);

return 0;
}

相关问题

高精加减乘除用c++写出来

当然可以，以下是一个简单的高精度加减乘除的 C 代码示例：

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

#define MAX_LEN 1000

typedef struct {
    int len;
    int num[MAX_LEN];
} BigInt;

void init(BigInt *a) {
    memset(a->num, 0, sizeof(a->num));
    a->len = 0;
}

void assign(BigInt *a, char *s) {
    init(a);
    int len = strlen(s);
    for (int i = len - 1, j = 0, k = 1; i >= 0; i--) {
        a->num[j] += (s[i] - '0') * k;
        if (k == 10000) {
            j++;
            k = 1;
        } else {
            k *= 10;
        }
    }
    a->len = len / 4 + 1;
    while (a->len > 1 && a->num[a->len - 1] == 0) {
        a->len--;
    }
}

void print(BigInt *a) {
    printf("%d", a->num[a->len - 1]);
    for (int i = a->len - 2; i >= 0; i--) {
        printf("%04d", a->num[i]);
    }
    printf("\n");
}

int compare(BigInt *a, BigInt *b) {
    if (a->len != b->len) {
        return a->len - b->len;
    }
    for (int i = a->len - 1; i >= 0; i--) {
        if (a->num[i] != b->num[i]) {
            return a->num[i] - b->num[i];
        }
    }
    return 0;
}

void add(BigInt *a, BigInt *b, BigInt *c) {
    init(c);
    int carry = 0;
    for (int i = 0; i < a->len || i < b->len; i++) {
        int sum = a->num[i] + b->num[i] + carry;
        c->num[c->len++] = sum % 10000;
        carry = sum / 10000;
    }
    if (carry > 0) {
        c->num[c->len++] = carry;
    }
}

void sub(BigInt *a, BigInt *b, BigInt *c) {
    init(c);
    int borrow = 0;
    for (int i = 0; i < a->len || i < b->len; i++) {
        int diff = a->num[i] - b->num[i] - borrow;
        if (diff < 0) {
            diff += 10000;
            borrow = 1;
        } else {
            borrow = 0;
        }
        c->num[c->len++] = diff;
    }
    while (c->len > 1 && c->num[c->len - 1] == 0) {
        c->len--;
    }
}

void mul(BigInt *a, BigInt *b, BigInt *c) {
    init(c);
    for (int i = 0; i < a->len; i++) {
        int carry = 0;
        for (int j = 0; j < b->len; j++) {
            int sum = a->num[i] * b->num[j] + c->num[i + j] + carry;
            c->num[i + j] = sum % 10000;
            carry = sum / 10000;
        }
        if (carry > 0) {
            c->num[i + b->len] += carry;
        }
    }
    c->len = a->len + b->len;
    while (c->len > 1 && c->num[c->len - 1] == 0) {
        c->len--;
    }
}

void div(BigInt *a, BigInt *b, BigInt *c, BigInt *d) {
    init(c);
    init(d);
    for (int i = a->len - 1; i >= 0; i--) {
        d->num[0] = a->num[i];
        d->len = 1;
        int left = 0, right = 9999, mid;
        while (left <= right) {
            mid = (left + right) / 2;
            BigInt tmp, mul;
            init(&tmp);
            init(&mul);
            tmp.num[0] = mid;
            tmp.len = 1;
            mul(&tmp, b, &mul);
            if (compare(&mul, d) <= 0) {
                left = mid + 1;
            } else {
                right = mid - 1;
            }
        }
        c->num[i] = right;
        BigInt tmp, mul, sub;
        init(&tmp);
        init(&mul);
        init(&sub);
        tmp.num[0] = right;
        tmp.len = 1;
        mul(&tmp, b, &mul);
        sub(a, &mul, &sub);
        assign(a, "");
        add(&sub, &tmp, a);
    }
    while (c->len > 1 && c->num[c->len - 1] == 0) {
        c->len--;
    }
}

int main() {
    BigInt a, b, c, d;
    char s1[MAX_LEN], s2[MAX_LEN];
    scanf("%s%s", s1, s2);
    assign(&a, s1);
    assign(&b, s2);
    add(&a, &b, &c);
    sub(&a, &b, &c);
    mul(&a, &b, &c);
    div(&a, &b, &c, &d);
    print(&c);
    print(&d);
    return 0;
}

这个代码实现了高精度加减乘除四个基本运算，可以处理超过 long long 范围的整数。