两个极大正整数求和C++源程序

#include #include using namespace std;int main() string a, b;int *sum;int i, ja, jb;cout 请输入两个数字，每个数字以空格结尾： a b;string temp;if ( a.size() b.size() ) / 如果a长度大于b，则交换两字符串temp = a; a = b;b = temp;/ 在这后，就肯定是a的长度比b小，所以和的位数最大就是b的位数+1。sum = new intb.size()+1; / 所以给和申请b的位数+1个位置。sum0 = 0;char achar1;char bchar1;int carry = 0; / 记录进位int sizeDif = b.size() - a.size(); / 记录两个数字的位数差int count; / 记录和的位数for ( ja = a.size()-1, jb = b.size()-1; ja = 0; ja-, jb- )achar0 = aja;bchar0 = bjb;sumjb+1 = atoi(achar) + atoi(bchar) + carry;carry = 0; / 清空进位标志位if( sumjb+1 9 )carry = 1;sumjb+1 -= 10; / 使之成为一位数，方便输出。for ( i = sizeDif-1; i = 0 ; i- )bchar0 = bi;sumi+1 = atoi(bchar)+carry; / 转换为数字carry = 0; / 清空进位标志位if( sumi+1 9 )carry = 1;sumi+1 -= 10; / 使之成为一位数，方便输出。cout endl The process 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 ) / 当和的位数为b的位数+1时cout 1; / 先输出最高位，否则就不输出。elsecout ;for ( i = 1; i = b.size(); i+ )cout sumi;cout endl;cout endl So the result is: ;int remainder;if ( carry = 1 ) / 当和的位数为b的位数+1时count = b.size()+1;sum0 = 1;remainder = count%4;for ( i = 0; i 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 = co