D:大整数的加减乘除
2024-08-30 23:06:43
描述
给出两个正整数以及四则运算操作符(+ - * /),求运算结果。
输入第一行:正整数a,长度不超过100
第二行:四则运算符o,o是“+”,“-”,“*”,“/”中的某一个
第三行:正整数b,长度不超过100
保证输入不含多余的空格或其它字符输出一行:表达式“a o b”的值。
补充说明:
1. 减法结果有可能为负数
2. 除法结果向下取整
3. 输出符合日常书写习惯,不能有多余的0、空格或其它字符样例输入
9876543210
+
9876543210
样例输出
19753086420
Code:
#include<iostream>
#include<cstring>
using namespace std; const int MAXLEN = 201; int Substract(int *p1, int *p2, int len1, int len2) {
int i;
if (len1 == len2) {
for (i = len1 - 1; i >= 0; ++i) {
if (p1[i] < p2[i]) return -1; // p1 < p2
else if (p1[i] > p2[i]) break; // p1 > p2
}
}
for (i = 0; i < len1; ++i) { // 要求调用本函数确保当i >= len2 时, p2[i] = 0
p1[i] -= p2[i];
if (p1[i] < 0) {
p1[i] += 10;
--p1[i+1];
}
}
for (i = len1 - 1; i >= 0; --i) {
if (p1[i]) return i + 1; // 找到最高位第一个不为0
return 0; // 全部为0说明两者相等
}
} class Integer {
private:
int is_neg;
int len;
int s[MAXLEN];
char str[MAXLEN]; public:
Integer(const char *string = "") {
memset(s, 0, MAXLEN*sizeof(int));
memset(str, 0, MAXLEN*sizeof(char));
strcpy(str, string);
is_neg = 0;
len = strlen(str);
for (int i = 0; i < len; ++i) {
s[i] = int(str[len-1-i]) - 48; // s[i]低位在左, 高位在右。
}
} Integer& operator = (const Integer& oth) {
if (this == &oth) return *this;
memset(s, 0, MAXLEN*sizeof(int));
memset(str, 0, MAXLEN*sizeof(char));
is_neg = oth.is_neg;
len = oth.len;
for (int i = 0; i < oth.len; ++i) {
s[i] = oth.s[i];
}
strcpy(str, oth.str);
return *this;
} bool operator == (const Integer& oth) {
if (this == &oth) return true;
bool ret = true;
if (len != oth.len || is_neg != oth.is_neg)
ret = false;
if (strcmp(str, oth.str)) ret = false;
for (int i = 0; i < oth.len; ++i) {
if (s[i] != oth.s[i]) ret = false;
}
return ret;
} bool operator != (const Integer& oth) {
return !(*this == oth);
} Integer operator+(const Integer& oth) {
Integer c;
int length = len >= oth.len ? len : oth.len;
length += 1;
c.len = length;
for (int i = 0; i < c.len; ++i) {
c.s[i] += s[i] + oth.s[i];
c.s[i+1] += c.s[i] / 10;
c.s[i] = c.s[i] % 10;
} while ((c.len > 1) && (c.s[c.len-1] == 0))
c.len--; return c;
} Integer operator-(const Integer& oth) {
Integer c;
int flag = 0; if (len > oth.len) flag = 1;
else if (len < oth.len) flag = -1;
else flag = strcmp(str, oth.str); if (flag >= 0) {
c.len = len;
int borrow = 0;
for (int i = 0; i < c.len; ++i) {
c.s[i] += s[i] - oth.s[i];
if (borrow) c.s[i] -= 1;
if (c.s[i] < 0) {
c.s[i] += 10;
borrow = 1;
} else {
borrow = 0;
}
}
while ((c.len > 1) && (c.s[c.len-1] == 0))
c.len--;
} else {
c.is_neg = 1;
c.len = oth.len;
int borrow = 0;
for (int i = 0; i < c.len; ++i) {
c.s[i] += oth.s[i] - s[i];
if (borrow) c.s[i] -= 1;
if (c.s[i] < 0) {
c.s[i] += 10;
borrow = 1;
} else {
borrow = 0;
}
}
while ((c.len > 1) && (c.s[len-1] == 0))
c.len--;
}
return c;
} Integer operator*(const Integer& oth) {
Integer c;
c.len = len + oth.len + 1;
for (int i = 0; i < len; ++i) {
for (int j = 0; j < oth.len; ++j) {
c.s[i+j] += s[i] * oth.s[j];
c.s[i+j+1] += c.s[i+j] / 10;
c.s[i+j] = c.s[i+j] % 10;
}
}
while ((c.len > 1) && (c.s[c.len-1] == 0))
c.len--; return c;
} Integer operator/(Integer& oth) {
Integer c;
c.len = MAXLEN;
int i, temp;
if (len < oth.len) {
for (int i = 0; i < MAXLEN; ++i) {
c.s[i] = 0;
}
return c;
}
int nTimes = len - oth.len;
if (nTimes > 0) {
for (i = len-1; i >= nTimes; --i) {
oth.s[i] = oth.s[i-nTimes];
}
for (; i >= 0; --i) {
oth.s[i] = 0;
}
oth.len = len;
}
for (i = 0; i <= nTimes; ++i) {
while ((temp = Substract(s, oth.s+i, len, oth.len-i)) >= 0) {
len = temp;
++c.s[nTimes-i];
}
}
while ((c.len > 1) && (c.s[c.len-1] == 0))
c.len--; return c;
} friend ostream & operator<<(ostream& out, const Integer& oth) {
if (oth.is_neg) out << "-";
for (int k = oth.len-1; k >= 0; --k)
out << oth.s[k];
return out;
}
}; int main() {
char s1[MAXLEN], s2[MAXLEN];
char ope;
cin >> s1;
cin >> ope;
cin >> s2;
Integer a(s1);
Integer b(s2); switch (ope) {
case '+':
cout << a + b << endl;
break;
case '-':
cout << a - b << endl;
break;
case '*':
cout << a * b << endl;
break;
case '/':
cout << a / b << endl;
break;
} return 0;
}
reference:
https://blog.csdn.net/ApplePeel_90/article/details/80066215
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