C ++中两个大数的和
在这个问题中，我们得到了两个定义两个大数字的字符串。我们的任务是创建一个程序来查找两个大数的和。
让我们举个例子来了解这个问题，
Input: number1 = “341299123919” number2 = “52413424” Output: 341351537343
为了解决这个问题，我们将遍历这两个字符串。并逐位相加并传播进位。并逐位存储结果以求和。
算法
Initialize sum = 0, carry = 0. Step 1: loop from n to 0. Step 1.1: intSum = number1[i] + number2[i] Step 1.2: carry = intSum/10. Sum += intSum Step 2: sum += carry. Step 3: return sum.
示例
该程序说明了我们解决方案的工作原理，
#include<bits/stdc++.h>
using namespace std;
string addBigNumbers(string number1, string number2) {
if (number1.length() > number2.length())
swap(number1, number2);
string sum = "";
int len1 = number1.length();
int len2 = number2.length();
int digitDiff = len2 - len1;
int carry = 0;
int intSum;
for (int i=len1-1; i>=0; i--) {
intSum = ((number1[i]-'0') + (number2[i+digitDiff]- '0') + carry);
sum.push_back(intSum%10 + '0');
carry = intSum/10;
}
for (int i=digitDiff-1; i>=0; i--) {
intSum = ((number2[i]-'0')+carry);
sum.push_back(intSum%10 + '0');
carry = intSum/10;
}
if (carry)
sum.push_back(carry+'0');
reverse(sum.begin(), sum.end());
return sum;
}
int main() {
string number1 = "235235823852";
string number2 = "45230820348";
cout<<"Sum of two large numbers is "<<addBigNumbers(number1,x number2);
return 0;
}输出结果
Sum of two large numbers is 280466644200