兩個極大正整數(shù)求和C++源程序

1、#include <iostream>#include <string>using namespace std;int main() string a, b;int *sum;int i, ja, jb;cout << "請輸入兩個數(shù)字，每個數(shù)字以空格結尾：" << endl;cin >> a >> b;string temp;if ( a.size() > b.size() ) / 如果a長度大于b，則交換兩字符串temp = a; a = b;b = temp;/ 在這后，就肯定是a的長度比b

2、小，所以和的位數(shù)最大就是b的位數(shù)+1。sum = new intb.size()+1; / 所以給和申請b的位數(shù)+1個位置。sum0 = 0;char achar1;char bchar1;int carry = 0; / 記錄進位int sizeDif = b.size() - a.size(); / 記錄兩個數(shù)字的位數(shù)差int count; / 記錄和的位數(shù)for ( ja = a.size()-1, jb = b.size()-1; ja >= 0; ja-, jb- )achar0 = aja;bchar0 = bjb;sumjb+1 = atoi(achar) + atoi(b

3、char) + carry;carry = 0; / 清空進位標志位if( sumjb+1 > 9 )carry = 1;sumjb+1 -= 10; / 使之成為一位數(shù)，方便輸出。for ( i = sizeDif-1; i >= 0 ; i- )bchar0 = bi;sumi+1 = atoi(bchar)+carry; / 轉換為數(shù)字carry = 0; / 清空進位標志位if( sumi+1 > 9 )carry = 1;sumi+1 -= 10; / 使之成為一位數(shù)，方便輸出。cout << endl << "The proces

4、s is:" << endl;for ( i = 0; i <= sizeDif; i+ )cout << " "cout << a << endl;cout << "+" << b << endl;for ( i = 0; i <= b.size(); i+ )cout << "-"cout << endl;if ( carry = 1 ) / 當和的位數(shù)為b的位數(shù)+1時cout << 

5、9;1' / 先輸出最高位，否則就不輸出。elsecout << " "for ( i = 1; i <= b.size(); i+ )cout << sumi;cout << endl;cout << endl << "So the result is: "int remainder;if ( carry = 1 ) / 當和的位數(shù)為b的位數(shù)+1時count = b.size()+1;sum0 = 1;remainder = count%4;for ( i = 0; i <

6、 remainder; i+ )cout << sumi;if ( remainder = 0 )for ( ; remainder < 4; remainder+ )cout << sumremainder;for ( i = remainder; i < count; i+ )if ( (i-remainder) % 4 = 0)cout << ','cout << sumi;else count = b.size();remainder = count%4;for ( i = 1; i < remainder+1; i+ )cout << sumi;for ( i = remainder+1; i <= count; i+ )if (