题解 | #浮点数加法#

#include <iostream>
#include <cstdio>
#include <string>
using namespace std;

int main() {
    char str1[101], str2[101];
    while (scanf("%s%s", str1, str2) != EOF) {
        string string1 = str1;
        string string2 = str2;
        string ret;
        int length1 = string1.size() - 1;
        int length2 = string2.size() - 1;
        int point1 = string1.find('.');
        int point2 = string2.find('.'); //以小数点处作为中轴线
        int decade = 0;
        while (length1 >= 0 || length2 >= 0) {
            int judge1 = length1 >= 0 ? (point1 > length1 ? point1 - length1 : length1 -
                                         point1) : -1;//记录当前位距离小数点的距离
            int judge2 = length2 >= 0 ? (point2 > length2 ? point2 - length2 : length2 -
                                         point2) : -1;//如果str遍历完，置负接下来不参与计算
            if (judge1 < judge2) {//当字符串位数不对应时
                int num = string2[length2--] - '0' + decade;
                int remain = num % 10;
                decade = num / 10;
                ret = char(remain + '0') + ret;
            } else if (judge1 > judge2) {
                int num = string1[length1--] - '0' + decade;
                int remain = num % 10;
                decade = num / 10;
                ret = char(remain + '0') + ret;
            } else if (judge1 == judge2 &&
                       string1[length1] != '.') {//当字符串位数对应时
                int num = string1[length1--] + string2[length2--] - '0' - '0' + decade;
                int remain = num % 10;
                decade = num / 10;
                ret = char(remain + '0') + ret;
            } else if (string1[length1] == '.') {
                ret = '.' + ret;
                length1--;
                length2--;
            }
        }
        printf("%s\n", ret.c_str());
    }
}
// 64 位输出请用 printf("%lld")

笨方法，用小数点对齐后再按位数依次相加