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  • Codeforces Round #339 (Div. 2) A

    Description

    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).

    Sample Input

    1 10 2

    2  4  5

    Sample Output

    1 2 4 8

    -1

    此题非常要注意什么时候循环结束,否则可能会爆long long

    #include<stdio.h>
    //#include<bits/stdc++.h>
    #include<string.h>
    #include<iostream>
    #include<math.h>
    #include<sstream>
    #include<set>
    #include<queue>
    #include<vector>
    #include<algorithm>
    #include<limits.h>
    #define inf 0x3fffffff
    #define lson l,m,rt<<1
    #define rson m+1,r,rt<<1|1
    #define LL long long
    using namespace std;
    LL powll(LL x, LL n)
    {
        LL pw = 1;
        while (n > 0)
        {
            if (n & 1)      
                pw *= x;
            x *= x;
            n >>= 1;      
        }
        return pw;
    }
    int main()
    {
       LL l,r,k;
       LL x;
       cin>>l>>r>>k;
       int flag=0;
       for(int i=0;i<=66;i++)
       {
           if(x>r/k) break;
           x=powll(k,i);
           if(x>=l&&x<=r)
           {
               flag=1;
               cout<<x<<" ";
           }
       }
       if(flag==0)
       {
           puts("-1");
       }
       return 0;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/yinghualuowu/p/5134584.html
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