做网站生成二维码苏州新闻今天最新消息新闻事件

题目：假设输入是2个数字，可能超过long long类型能表示的范围，请输出两数相加的运算结果。

思路：2个数输入的时候，肯定都是用string存的，先将短的数在末尾补0，使得二者一样长。然后挨个位相加，并设置一个标志位专门用来存溢出(即进位)情况。
代码如下：

#include <stdio.h>
#include <string>using namespace std;static void compasent0(std::string& inputStr1, std::string& inputStr2)
{if (inputStr1.size() < inputStr2.size()){int diff = inputStr2.size() - inputStr1.size();string prefix(diff, '0');inputStr1 = prefix + inputStr1;}else{{int diff = inputStr1.size() - inputStr2.size();string prefix(diff, '0');inputStr2 = prefix + inputStr2;}}
}static bool checkStrValid(const std::string inputStr)
{for(const char& ch : inputStr){if ((ch < '0') || (ch > '9')){return false;}}return true;
}static std::string BigNumAdd(std::string inputStr1, std::string inputStr2)
{if (!checkStrValid(inputStr1) || !checkStrValid(inputStr2)){printf("ERROR, invalid input!\n");return "ERROR";}compasent0(inputStr1, inputStr2);printf("line[%u], inputStr1[%s], inputStr2[%s]\n", __LINE__, inputStr1.c_str(), inputStr2.c_str());std::string out(inputStr1.size(), '0');int tempSum = 0;int overflow = 0;for (int i = inputStr1.size() - 1; i > -1; --i){tempSum = (int)(inputStr1[i] - '0') + (int)(inputStr2[i] - '0') + overflow;if (tempSum < 10){out[i] = (char)(tempSum + '0');overflow = 0;}else{out[i] = (char)(tempSum % 10 + '0');overflow = 1;}}if (overflow){out = to_string(overflow) + out;}return out;
}int main()
{std::string inputStr1 = "1906";std::string inputStr2 = "456";std::string outputStr = BigNumAdd(inputStr1, inputStr2);printf("outputStr[%s]\n", outputStr.c_str());inputStr1 = "99999999999999999999";inputStr2 = "1";outputStr = BigNumAdd(inputStr1, inputStr2);printf("outputStr[%s]\n", outputStr.c_str());return 0;
}

输出是：

./a.out
line[45], inputStr1[1906], inputStr2[0456]
outputStr[2362]
line[45], inputStr1[99999999999999999999], inputStr2[00000000000000000001]
outputStr[100000000000000000000]