zzulioj - 2624: 小H的奇怪加法(模拟)

题目链接：http://acm.zzuli.edu.cn/problem.php?id=2624 

　　模拟加法进位就行了，写起来和高精度加法类似。

#include<set>
#include<map>
#include<stack>
#include<queue>
#include<cmath>
#include<cstdio>
#include<cctype>
#include<string>
#include<vector>
#include<climits>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#define max(a, b) (a > b ? a : b)
#define min(a, b) (a < b ? a : b)
#define mst(a) memset(a, 0, sizeof(a))
#define _test printf("~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\n")
using namespace std;
typedef long long ll;
typedef pair<int, int> P;
const double eps = 1e-7;
const int INF = 0x3f3f3f3f;
const ll ll_INF = 0x3f3f3f3f3f3f3f;
const int maxn = 1e3+10;
char ls[20];
char num1[maxn], num2[maxn], res[maxn];
void add() { //高精度整数加法(改)
    mst(res);
    int len1 = strlen(num1);
    int len2 = strlen(num2);
    int len3 = strlen(ls);
    reverse(num1, num1+len1);
    reverse(num2, num2+len2);
    int len, carry;
    len = carry = 0;
    for (int i = 0; i<len1 || i<len2; ++i) {
        carry += i<len1 ? num1[i] - '0' : 0;
        carry += i<len2 ? num2[i] - '0' : 0;
        if (ls[len3-i-1] != '0') { //非0即为2~9进制
            res[len] = carry%(ls[len3-i-1]-'0') + '0';
            carry /= (ls[len3-i-1] - '0');
        }
        else { //是0就按10进制
            res[len] = carry%10 + '0';
            carry /= 10;
        }
        ++len;
    }
    if (carry) //判断最后一位是不是有进位
        res[len++] = carry + '0';
    res[len] = 0;
    for (int i = len-1; i>0; --i) { //排除先导0的影响 注意 不能直接i>=0， 如果结果刚好等于0就会出错了
        if (res[i] == '0')
            res[i] = 0;
        else 
            break;
    }
    reverse(res, res+strlen(res));
}   
int main(void) {
    scanf("%s %s %s", ls, num1, num2);
    add();
    printf("%s\n", res);
    return 0;
}