题解 | #浮点数加法#

#include <stdio.h>
// 计算字符串长度：
int Length(char a[]) {
    int n = 0;
    for (int i = 0; a[i] != '\0'; i++) {
        n++;
    }
    return n;
}
int main() {
    char a[100] = {'0'};
    char b[100] = {'0'};
    char c[100] = {'0'};
    scanf("%s", a);
    scanf("%s", b);
    int len1 = Length(a);
    int len2 = Length(b);
    int i = 0, j = 0, k = 0;
    while (a[i] != '.') {
        i++;
    }
    while (b[j] != '.') {
        j++;
    }//判断整数位位数是否相同
    if (i > j) {//第一组输入的数整数位长，将第二组前面补0到和第一组整数位位数相等
        char d[100] = "";
        int m;
        for (m = 0; m < i - j; m++) {
            d[m] = '0';
        }
        for (int n = 0; n < len2; n++) {
            d[m] = b[n];
            m++;
        }
        int len4 = Length(d);
        for (int p = 0; p < len4; p++) {
            b[p] = d[p];
        }
    } else if (i < j) {//第二组输入的数整数位长，将第一组前面补0到和第二组整数位位数相等
        char d[100] = "";
        int m;
        for (m = 0; m < j - i; m++) {
            d[m] = '0';
        }
        for (int n = 0; n < len1; n++) {
            d[m] = a[n];
            m++;
        }
        int len4 = Length(d);
        for (int p = 0; p < len4; p++) {
            a[p] = d[p];
        }
    }
    len1 = Length(a);
    len2 = Length(b);
    if (len1 > len2) {//将小数位短的补0到和小数位长的位数相等
        k = len1 - 1;
        for (int l = len2; l < len1; l++) {
            b[l] = '0';
        }
    } else {
        k = len2 - 1;
        for (int l = len1; l < len2; l++) {
            a[l] = '0';
        }
    }
    for(int q=0;q<=k;q++)
    {
        c[q]='0';
    }
    while (k != -1) {//从后往前逐位相加，满十进一
        if (a[k] == '.') {
            c[k] = '.';
            k--;
        } else {
            if ((a[k] - '0') + (b[k] - '0') +(c[k]-'0')> 9) {
                if (a[k - 1] == '.') {
                    c[k - 2] = '1';
                    c[k] = '0' + ((a[k] - '0') + (b[k] - '0') + (c[k] - '0') - 10);
                } else {
                    c[k - 1] = '1';
                    c[k] = '0' + ((a[k] - '0') + (b[k] - '0') +(c[k] - '0') - 10);
                }
            } else {
                c[k] = '0' + (a[k] - '0') + (b[k] - '0')+(c[k] - '0');
            }
            k--;
        }
    }
    printf("%s\n", c);
    return 0;
}