hdu 1333水题

刚看的时候还以为挺难的，试了一下最水的方法，居然过了。就是从N以后往下试，试到smith数停止就行，而且对于smith数的判断也可以简单处理，循环到sqrt(n)即可，唉，这种简单方法我这次现场赛咋没想到呢……

/*
 * hdu1333/win.cpp
 * Created on: 2012-10-27
 * Author    : ben
 */
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <queue>
#include <set>
#include <map>
#include <stack>
#include <string>
#include <vector>
#include <deque>
#include <list>
#include <functional>
#include <numeric>
#include <cctype>
using namespace std;
int getbitsum(int n) {
    int ret = 0;
    while(n > 0) {
        ret += n % 10;
        n /= 10;
    }
    return ret;
}
bool isSmith(int n) {
    int sum1 = getbitsum(n);
    int sum2 = 0;
    int sqrtn = (int)sqrt(n);
    bool isprime = true;
    for(int i = 2; i <= sqrtn; i++) {
        while(n % i == 0) {
            isprime = false;
            sum2 += getbitsum(i);
            n /= i;
        }
        if(n == 1) {
            break;
        }
    }
    if(n > 1) {
        sum2 += getbitsum(n);
    }
    return sum1 == sum2 && !isprime;
}

int main() {
#ifndef ONLINE_JUDGE
    freopen("data.in", "r", stdin);
#endif
    int n;
    while(scanf("%d", &n) == 1 && n > 0) {
        do{
            n++;
        }while(!isSmith(n));
        printf("%d\n", n);
    }
    return 0;
}