题解 | #大整数的因子#

//除法的原理，KY26和KY30都用到了，如果上述两题会，这道题还是比较简单的
#include "stdio.h"
#include "string"
using namespace std;
char num[31];string str;
string divsionJudge(){
    str = num;int k = 2;//为除数
    int length = str.size();
    char shadow[31];//作为num[31]的影子寄存器
    int reminder = 0;str = "";int t;//remainder为余数，str记录能整除的k
    bool flag = false;//记录有没有能整除的数
    while (k<=9){
        for (int i = 0; i < length; ++i) {//每次整除判断时，都对影子寄存器初始化
            shadow[i] = num[i];
        }
        reminder = 0;int t;
        for (int i = 0; i < length; ++i) {
            t = (reminder*10 + shadow[i] - '0')%k;
            shadow[i] = (reminder*10 + shadow[i] - '0')/k + '0';
            reminder = t;
        }
        if (reminder == 0){
            flag = true;
            str += char (k + '0');
        }
        ++k;
    }
    if (flag)
        return str;
    else
        return "none";
}

void Init(){//对num和str初始化
    for (int i = 0; i < 31; ++i) {
        num[i]='\0';
    }
    str.clear();
}

int main(){
    while (scanf("%s",num)!=EOF){
        if (num[0] == '-')
            break;
        if (num[0] == '1' && num[1] == '\0'){
            printf("none\n");
        } else if (num[0] == '0' && num[1] == '\0'){
            printf("2 3 4 5 6 7 8 9\n");
        } else{
            string result = divsionJudge();
            if (result != "none"){
                for (int i = 0; i < result.size()-1; ++i) {
                    printf("%c ",result[i]);
                }
                printf("%c\n",result[result.size()-1]);
                Init();
            } else{
                printf("%s\n",result.c_str());
                Init();
            }
        }
    }
}