CODEFORCES #339 div2 A

A. Link/Cut Tree

Programmer Rostislav got seriously interested in the Link/Cut Tree data structure, which is based on Splay trees. Specifically, he is now studying the expose procedure.

Unfortunately, Rostislav is unable to understand the definition of this procedure, so he decided to ask programmer Serezha to help him. Serezha agreed to help if Rostislav solves a simple task (and if he doesn't, then why would he need Splay trees anyway?)

Given integers l, r and k, you need to print all powers of number k within range from l to r inclusive. However, Rostislav doesn't want to spent time doing this, as he got interested in playing a network game called Agar with Gleb. Help him!

Input

The first line of the input contains three space-separated integers l, r and k (1 ≤ l ≤ r ≤ 1018, 2 ≤ k ≤ 109).

Output

Print all powers of number k, that lie within range from l to r in the increasing order. If there are no such numbers, print "-1" (without the quotes).

题目简单来说就是输出在区间[l,r]上所有k^x。水题，但是要留意l,r的最大值可为10^18，k的最大值可达到10^9若不留意比较方法可能会造成数据溢出而导致结果出错。

所以我在做的时候决定用tem存储当前可能会输出的数，将r/tem与k比较以此来决定是否输出结束，可以防止数据溢出。

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;

int main()
{
long long l,r,k;
while (scanf("%I64d%I64d%I64d",&l,&r,&k)==3)
{
long long tem=1;
int cnt=0;
if (k==1)
{
if (l<=1&&r>=1)
printf("%I64d\n",k);
else
printf("-1\n");
}
else
{
if (tem<l)
{
while (l/tem>=k)
tem*=k;
}
if (tem<l&&r/tem<k)
printf("-1\n");
else
{
if (tem<l)
tem*=k;
printf("%I64d",tem);
while (r/tem>=k)
{
tem*=k;
printf(" %I64d",tem);
}
printf("\n");
}
}
}
return 0;
}