二分走一波，没想到题解的大佬做法 p_q

注意爆long long，所以先对数取一下上限

二分确定下限，然后输出

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stack>
#include<set>
#include<queue>
#include<vector>
#include<iostream>
#include <limits.h>
#include<algorithm>
#define MAXN 1010100
#define LL unsigned long long
// #define ll __int64
#define INF 0x7fffffff
#define cs(s) freopen(s,"r",stdin)
#define mem(x) memset(x,0,sizeof(x))
#define PI acos(-1)
#define eps 1e-10
using namespace std;
int gcd(int a,int b){return b?gcd(b,a%b):a;}
int lcm(int a,int b){return a/gcd(a,b)*b;}
LL powmod(LL a,LL b){LL ans=1;while(b){if(b%2)ans=ans*a;a=a*a;b/=2;}return ans;}
//head
LL ll,rr,w,L,R,cnt;
LL mid,ans;
int main(){
    scanf("%llu%llu%llu",&ll,&rr,&w);
    if(w==1){
        puts("1");
        return 0;
    }
    int l=0,r=log(rr)/log(w);
    R=r;
    l=0,r=log(rr)/log(w);
    ans=-1;
    while(l<=r){
        mid=(l+r)/2;
        if(powmod(w,mid)>=ll)ans=mid,r=mid-1;
        else l=mid+1;
    }
    L=ans;
    for(int i=L;i<=R;i++){
        ans=powmod(w,i);
        if(ans>=ll&&ans<=rr)
        printf("%lld ",ans),cnt++;
    }
    if(!cnt)puts("-1");
    return 0;
}