实现两个数的加法操作，不考虑 负数的情况
#include
#include
#include
using namespace std;
string NumAdd(const string &n1, const string &n2)
{
int len1 = n1.length();
int len2 = n2.length();
std::string num = "0123456789";
int f = 0;
int k = 0;
int i, j;
std::string sum = "";
for (i = len1 - 1, j = len2 - 1; i >= 0 && j >= 0; i--, j--)
{
int s = (n1[i] - '0') + (n2[j] - '0') + f;
f = s / 10;
k = s % 10;
sum += num[k];
}
for (; i >= 0; i--) {
int s = (n1[i] - '0') + f;
f = s / 10;
k = s % 10;
sum += num[k];
}
for(; j >=0; j--) {
int s = (n2[j] - '0') + f;
f = s / 10;
k = s % 10;
sum += num[k];
}
if (f > 0) {
sum += num[f];
}
std::reverse(sum.begin(), sum.end());
return sum;
}
int main(int argc, char *argv[])
{
std::string n1 = "6234456711111111111112341234";
std::string n2 = "5678245111111111111111111111";
std::string sum = NumAdd(n1, n2);
cout << n1 << " + " << n2 << " = " << sum << endl;
return 0;
}
输出结果：
6234456711111111111112341234 + 5678245111111111111111111111 = 11912701822222222222223452345
