Codeforces Round #262 (Div. 2) B

B. Little Dima and Equation

        题意：输入a b c，求满足等式 x = b·s(x)^a + c 的所有x，s(x)表示x各位之和。

        思路：因为x的范围是1~999999999，那么s(x)的范围就是1~81这样的话枚举每一个s(x)可能的值，计算出b·s(x)^a + c，看是否满足等式。

#include <iostream>    #include <stdio.h>    #include <cmath>    #include <algorithm>    #include <iomanip>    #include <cstdlib>    #include <string>    #include <memory.h>    #include <vector>    #include <queue>    #include <stack>    #include <map>  #include <set>  #include <ctype.h>    #define INF 10000000#define ll long long#define min3(a,b,c) min(a,min(b,c))#define max3(a,b,c) max(a,max(b,c))#define MAXN 100010using namespace std;  ll tmp[90];inline ll sum(ll n){    ll re=0;    while(n){        re+=n%10;        n/=10;    }    return re;}inline ll apowb(ll a,ll b){     ll re=1;     while(b--){         re*=a;         }            return  re;      }int main(){    ll a,b,c;    while(cin>>a>>b>>c){        int cnt=0;        vector<ll> ans;        for(ll i=1;i<=81;i++){            tmp[i]=b*apowb(i,a)+c;            if(tmp[i]>999999999LL)continue;            if(sum(tmp[i])==i){                cnt++;                ans.push_back(tmp[i]);            }        }        cout<<cnt<<endl;for(int i=0;i<ans.size();i++){            cout<<ans[i]<<" ";        }    }    return 0;}