解法一:
#include <iostream>
#include <string>
using namespace std;
int main(){
int n, m = 0; //n为除数,m为商
string str; //str为被除数
cin >> str >> n;
//被数为一位数且被除数小于除数的情况
if (str.length() == 1 && str[0] - '0'<n)
{
cout << "0 " << str[0] - '0' << endl;
}
else
{
for (int i = 0; i<str.length(); i++)
{
if (i>0 && m == 0 && str[i] - '0'<n)
cout << "0";
m = m * 10 + str[i] - '0';
if (m / n>0){
cout << m / n;
m = m % n;
}
}
cout << " " << m << endl;
}
return 0;
}
解法二:
#include <iostream>
#include <string>
using namespace std;
int main(){
string A,Q="";
int B,R=0;
cin>>A>>B;
for(char a:A){//大整数除法
R=R*10+a-'0';
Q+=R/B+'0';
R%=B;
}
while(!Q.empty()&&Q.front()=='0')//删除Q首部的'0'
Q.erase(Q.begin());
printf("%s %d",Q.size()==0?"0":Q.c_str(),R);//Q为空先输出一个0再输出R
return 0;
}