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大整数的加减乘除运算

作者:互联网

#include<cstdio>
#include<cstring>

int r = 0;

struct bign {
    int d[1000];
    int len;

    bign() {//构造函数
        memset(d, 0, sizeof(d));
        len = 0;
    }
};

bign change(char str[]) {
    bign a;
    a.len = strlen(str);
    for (int i = 0; i < a.len; i++) {
        a.d[i] = str[a.len - 1 - i] - '0';
    }
    return a;
}

bign add(bign a, bign b) {
    bign c;
    int carry = 0;//标识进位
    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) {
    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) {
    bign c;
    int carry = 0;
    for (int i = 0; i < a.len; i++) {
        int temp = a.d[i] * b + carry;
        c.d[c.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一一对应,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;//不够除则该位为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;
}

void print(bign a) {
    for (int i = a.len - 1; i >= 0; i--) {
        printf("%d", a.d[i]);
    }
}

int main() {
    char str1[1000], str2[1000];
    int r;
    scanf("%s%s", str1, str2);
    bign a = change(str1);
    bign b = change(str2);
    print(divide(a, 3, r));
    return 0;
}

标签:10,运算,int,len,++,bign,整数,carry,加减乘除
来源: https://blog.csdn.net/weixin_43340821/article/details/114705902