问题描述
判断两个数以及相加或相乘的结果是否超出int类型值得范围
思路
套用刘汝佳的大数模板,结果遇到两个错误。
Runtime Error:开的数组太小了(不懂具体开多少),尽量开大。
Wrong Answer:开始计算前应当注意清除两个大数的前缀0(但是输出的时候应是原样输出)。
代码
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
using namespace std;
const int maxn=1000;
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 int num)
{
char s[maxn];
sprintf(s,"%d",num);
*this=s;
return *this;
}
bign operator = (const char *num)
{
len=strlen(num);
for(int i=0;i<len;i++)
s[i]=num[len-i-1]-'0';
return *this;
}
//输出
string str() const
{
string res="";
for(int i=0;i<len;i++)
{
res=(char)(s[i]+'0')+res;
}
if(res=="")
res="0";
return res;
}
void clean()
{
while(len>1&&!s[len-1])
len--;
}
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;
}
bign operator - (const bign& b) const
{
bign c;
c.len=0;
for(int i=0,g=0;i<len;i++)
{
int x=s[i]-g;
if(i<b.len) x-=b.s[i];
if(x>=0)
g=0;
else
{
g=1;
x+=10;
}
c.s[c.len++]=x;
}
c.clean();
return c;
}
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-1;i++)
{
c.s[i+1]+=c.s[i]/10;
c.s[i]%=10;
}
c.clean();
return c;
}
bign operator / (const bign &b)const
{
bign ret,cur=0;
ret.len=len;
for(int i=len-1;i>=0;i--)
{
cur=cur*10;
cur.s[0]=s[i];
while(cur>=b)
{
cur-=b;
ret.s[i]++;
}
}
ret.clean();
return ret;
}
bign operator %(const bign &b) const
{
bign c=*this/b;
return *this-c*b;
}
bool operator <(const bign& b) const
{
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) const
{
return b < *this;
}
bool operator <=(const bign& b) const
{
return !(b < *this);
}
bool operator >=(const bign& b) const
{
return !(*this <b);
}
bool operator ==(const bign& b) const
{
return !(b< *this)&&!(*this<b);
}
bool operator !=(const bign& b) const
{
return *this >b || *this <b;
}
bign operator +=(const bign& b)
{
*this= *this + b;
return *this;
}
bign operator -=(const bign& b)
{
*this= *this - b;
return *this;
}
bign operator *=(const bign& b)
{
*this= *this * b;
return *this;
}
bign operator /=(const bign& b)
{
*this= *this / b;
return *this;
}
bign operator %=(const bign& b)
{
*this= *this % b;
return *this;
}
};
istream& operator >> (istream &in, bign& x)
{
string s;
in >> s;
x = s.c_str();
return in;
}
ostream& operator << (ostream &out, const bign& x)
{
out << x.str();
return out;
}
int main()
{
bign a,b,ans,inf;
inf="2147483647"; //int最大值
string c;
while(cin>>a>>c>>b)
{
cout<<a<<" "<<c<<" "<<b<<endl;
a.clean();
b.clean();
if(a>inf)
cout<<"first number too big\n";
if(b>inf)
cout<<"second number too big\n";
if(c=="+")
ans=a+b;
else
ans=a*b;
if(ans>inf)
cout<<"result too big\n";
}
return 0;
}