pta-ch4-作业代码.rar

#include <iostream> #include <string> using namespace std; /* 请在这里填写答案 */ class BigInt { public: friend istream& operator>>(istream&, BigInt&); friend ostream& operator<<(ostream&, BigInt&); friend BigInt& operator+(BigInt&, BigInt&); private: string val; }; istream& operator>>(istream& in, BigInt& n) { in >> n.val; return in; } ostream& operator<<(ostream& out, BigInt& n) { out << n.val; return out; } void invert(string& s) { for(int i = 0; i < s.size() / 2; i++) { swap(s.at(i), s.at(s.size() - 1 - i)); } } BigInt& operator+(BigInt& a, BigInt& b) { static BigInt c; string sa, sb, sc; sa = a.val; sb = b.val; invert(sa); //反转sa invert(sb); int Len_sa = sa.size(); int Len_sb = sb.size(); int i = 0; int e = 0; //e表示进位 while(i < Len_sa && i < Len_sb) { int x = (sa.at(i) - '0') + (sb.at(i) - '0') + e; e = x / 10; //进位，如果x大于9，进位1 x = x % 10; //进位后的x sc.push_back(x + '0'); //注意，这儿不能直接写 sc.at(i) = x + '0', sc的空间大小此时是0 i++; } while(i < Len_sa) { int x = (sa.at(i) - '0') + e; e = x / 10; //进位，如果x大于9，进位1 x = x % 10; //进位后的x sc.push_back(x + '0'); i++; } while(i < Len_sb) { int x = (sb.at(i) - '0') + e; e = x / 10; //进位，如果x大于9，进位1 x = x % 10; //进位后的x sc.push_back(x + '0'); i++; } if(e == 1) //最后的进位，如果是1，回到字符串最后 { sc.push_back(e + '0'); } invert(sc); c.val = sc; return c; } int main(){ BigInt a, b, c; cin>>a>>b; c=a+b; cout<<a<<"+"<<b<<"="<<c<<endl; return 0; }