A + B Problem II

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本文介绍了一种处理两个大整数（小于1000位）相加的方法，并通过C++实现了一个BigInteger类来完成这一操作。该类支持输入输出、加法运算等功能，并能处理进位问题。

http://acm.hdu.edu.cn/showproblem.php?pid=1002

#include <iostream> #include <string> using namespace std; /************************************************************************/ /* 两个小于1000位的大整数相加 */ /************************************************************************/ class BigInteger { public: BigInteger(); BigInteger operator+ (BigInteger &right); friend ostream& operator<<(ostream &out, BigInteger &other); friend istream& operator>>(istream &in, BigInteger &other); public: char arr[1000]; int realLength; }; int main() { BigInteger a, b; int N; while ((cin>>N) && (N>=0) && (N<=20)) { for (int i=0; i<N; i++) { cin>>a>>b; printf("Case %d:/n", i+1); cout<<a<<" + "<<b<<" = "<<(a+b)<<endl<<endl; } } } BigInteger::BigInteger() { memset(arr, '0', 1000); realLength= 0; } istream& operator>>(istream &in, BigInteger &other) { string str; cin>>str; int len = str.length(); for (int i=0; i<len; i++) { other.arr[1000-len+i] = str[i]; } other.realLength = len; return in; } ostream& operator<<(ostream &out, BigInteger &other) { int len = other.realLength; for (int i=0; i<len; i++) { cout<<other.arr[1000-len+i]; } return out; } BigInteger BigInteger::operator +(BigInteger &right) { BigInteger temp; int len = ((this->realLength >= right.realLength) ? this->realLength : right.realLength); int jinwei = 0; for (int i = len-1; i>=0; i--) { int t = (this->arr[1000-len+i] - '0') + (right.arr[1000-len+i] - '0') + jinwei; if (t >= 10) { jinwei = 1; temp.arr[1000-len+i] = t - 10 + '0'; } else { temp.arr[1000-len+i] = t + '0'; jinwei = 0; } } if (1 == jinwei) { temp.realLength = len + 1; temp.arr[999-len] = 1 + '0'; } else { temp.realLength = len; } return temp; }