题目链接：http://codeforces.com/problemset/problem/460/B
Little Dima misbehaved during a math lesson a lot and the nasty teacher Mr. Pickles gave him the following problem as a punishment.
Find all integer solutions x (0?<?x?<?109) of the equation:
where a, b, c are some predetermined constant values and function s(x) determines the sum of all digits in the decimal representation of number x.
The teacher gives this problem to Dima for each lesson. He changes only the parameters of the equation: a, b, c. Dima got sick of getting bad marks and he asks you to help him solve this challenging problem.
The first line contains three space-separated integers: a,?b,?c (1?≤?a?≤?5; 1?≤?b?≤?10000; ?-?10000?≤?c?≤?10000).
Print integer n — the number of the solutions that you‘ve found. Next print n integers in the increasing order — the solutions of the given equation. Print only integer solutions that are larger than zero and strictly less than 109.
3 2 8
3 10 2008 13726
1 2 -18
0
2 2 -1
4 1 31 337 967
代码如下：
#include <cstdio>
#include <cmath>
#include <cstring>
#include <string>
#include <iostream>
#include <algorithm>
using namespace std;
typedef __int64 LL;
LL ss[10017];
LL F(LL num)
{
LL s = 0;
while(num)
{
s += num%10;
num/=10;
}
return s;
}
LL Fac(LL num, LL n)
{
LL s = 1;
for(LL i = 1; i <= n; i++)
{
s*=num;
}
return s;
}
int main()
{
LL a, b, c;
while(~scanf("%I64d%I64d%I64d",&a,&b,&c))
{
memset(ss,0,sizeof(ss));
LL l = 0, i;
for(i = 1; i <= 81; i++)//枚举
{
LL x = b*Fac(i,a)+c;
if(x >= 1000000000 || x <= 0)
continue;
LL ans = (int)F(x);
if(ans == i)
ss[l++] = x;
}
printf("%I64d\n",l);
for(i = 0; i < l; i++)
{
if(i == 0)
printf("%I64d",ss[i]);
else
printf(" %I64d",ss[i]);
}
printf("\n");
}
return 0;
}
Codeforces Round #262 (Div. 2) 460B. Little Dima and Equation（枚举）,布布扣,bubuko.com
Codeforces Round #262 (Div. 2) 460B. Little Dima and Equation（枚举）
原文：http://blog.csdn.net/u012860063/article/details/38735883