练习C++bign类高精度,这一题数据中含有前导0,必须在重载=号时处理一下,而且必须以char类型的数组输入,一开始我用bign类输入,会报错的,我WA了至少30多次。
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <string>
#include <algorithm>
#include <climits>
using namespace std;
const int MAXN = 40000;
const int INF = 2147483647;
char str1[MAXN], str2[MAXN];
struct bign
{
int len, s[MAXN];
bign ()
{
memset(s, 0, sizeof(s));
len = 1;
}
bign (int num) {*this = num;}
bign (const char *num) { *this = num ;}
bign operator = (const char* num)
{
for(int i = 0; num[i] == '0'; num++) ; //去前导0,很重要。
len = strlen(num);
for(int i = 0; i < len; i++)
{
s[i] = num[len-i-1] - '0';
}
return *this;
}
bign operator = (int num)
{
char s[MAXN];
sprintf(s, "%d", num);
*this = s;
return *this;
}
bign operator + (const bign& b) const
{
bign c;
c.len = 0;
for(int i = 0, g = 0; g || i < max(len, b.len); i++)
{
int x = g;
if(i < len)
x += s[i];
if(i < b.len)
x += b.s[i];
c.s[c.len++] = x % 10;
g = x / 10;
}
return c;
}
void clean()
{
while(len > 1 && !s[len-1]) len--;
}
bign operator * (const bign &b) const
{
bign c;
c.len = len + b.len;
for(int i = 0; i < len; i++)
{
for(int j = 0; j < b.len; j++)
{
c.s[i+j] += s[i] * b.s[j];
}
}
for(int i = 0; i < c.len; i++)
{
c.s[i+1] += c.s[i] / 10;
c.s[i] %= 10;
}
c.clean();
return c;
}
bign operator += (const bign &b)
{
*this = *this + b;
return *this;
}
bool operator < (const bign &b)
{
if(len != b.len) return len < b.len;
for(int i = len-1; i >= 0; i--)
{
if(s[i] != b.s[i]) return s[i] < b.s[i];
}
return false;
}
bool operator > (const bign &b)
{
if(len != b.len) return len > b.len;
for(int i = len-1; i >= 0; i--)
{
if(s[i] != b.s[i]) return s[i] > b.s[i];
}
return false;
}
bool operator == (const bign &b)
{
return !(*this > b) && !(*this < b);
}
string str() const
{
string res = "";
for(int i = 0; i < len; i++) res = char(s[i]+'0') + res;
if(s[len-1] == 0) res = "";
if(res == "") res = "0";
return res;
}
};
istream& operator >> (istream &in, bign &x)
{
string s;
in >> s;
x = s.c_str();
return in;
}
ostream& operator << (ostream &out, bign &x)
{
out << x.str();
return out;
}
int main()
{
char ope;
bign a, b;
while(~scanf("%s", str1))
{
scanf(" %c", &ope);
scanf("%s", str2);
printf("%s %c %s\n", str1, ope, str2);
a = str1, b = str2;
if(a > INF) printf("first number too big\n");
if(b > INF) printf("second number too big\n");
if(ope == '+' && a+b > INF) printf("result too big\n");
else if(ope == '*' && a*b > INF) printf("result too big\n");
}
return 0;
}