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EOJ 2019.2月赛 D. 进制转换

时间:2019-02-23 13:29:06      阅读:197      评论:0      收藏:0      [点我收藏+]

https://acm.ecnu.edu.cn/contest/140/problem/D/

题意

  求一个区间L,R中,在K进制表示下末尾恰有m个0的数字个数。

思路

  末尾有m个0,就表示的是K^m的倍数,基本容斥,就是ans = X / (K ^m ) - X / (K ^(m+1));

技术分享图片
#include <algorithm>
#include  <iterator>
#include  <iostream>
#include   <cstring>
#include   <cstdlib>
#include   <iomanip>
#include    <bitset>
#include    <cctype>
#include    <cstdio>
#include    <string>
#include    <vector>
#include     <stack>
#include     <cmath>
#include     <queue>
#include      <list>
#include       <map>
#include       <set>
#include   <cassert>

/*

⊂_ヽ
  \\ Λ_Λ  来了老弟
   \(‘?‘)
    > ⌒ヽ
   /   へ\
   /  / \\
   ? ノ   ヽ_つ
  / /
  / /|
 ( (ヽ
 | |、\
 | 丿 \ ⌒)
 | |  ) /
‘ノ )  L?

*/

using namespace std;
#define lson (l , mid , rt << 1)
#define rson (mid + 1 , r , rt << 1 | 1)
#define debug(x) cerr << #x << " = " << x << "\n";
#define pb push_back
#define pq priority_queue



typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
//typedef __int128 bll;
typedef pair<ll ,ll > pll;
typedef pair<int ,int > pii;
typedef pair<int,pii> p3;

//priority_queue<int> q;//这是一个大根堆q
//priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q
#define fi first
#define se second
//#define endl ‘\n‘

#define boost ios::sync_with_stdio(false);cin.tie(0)
#define rep(a, b, c) for(int a = (b); a <= (c); ++ a)
#define max3(a,b,c) max(max(a,b), c);
#define min3(a,b,c) min(min(a,b), c);


const ll oo = 1ll<<17;
const ll mos = 0x7FFFFFFF;  //2147483647
const ll nmos = 0x80000000;  //-2147483648
const int inf = 0x3f3f3f3f;
const ll inff = 0x3f3f3f3f3f3f3f3f; //18
const int mod = 1e9+7;
const double esp = 1e-8;
const double PI=acos(-1.0);
const double PHI=0.61803399;    //黄金分割点
const double tPHI=0.38196601;


template<typename T>
inline T read(T&x){
    x=0;int f=0;char ch=getchar();
    while (ch<0||ch>9) f|=(ch==-),ch=getchar();
    while (ch>=0&&ch<=9) x=x*10+ch-0,ch=getchar();
    return x=f?-x:x;
}

inline void cmax(int &x,int y){if(x<y)x=y;}
inline void cmax(ll &x,ll y){if(x<y)x=y;}
inline void cmin(int &x,int y){if(x>y)x=y;}
inline void cmin(ll &x,ll y){if(x>y)x=y;}

/*-----------------------showtime----------------------*/


            ll solve(int k,int m,ll x){
                ll a  = x, b = x;
                for(int i=1; i<=m; i++){
                    a = a / k;
                    b = b / k;
                }
                b = b / k;

                return a - b;
            }
int main(){
            int T;
            scanf("%d", &T);
            while(T--){
                ll l,r; int k,m;
                cin>>l>>r>>k>>m;
                cout<<solve(k,m,r) - solve(k,m,l-1)<<endl;
            }

            return 0;
}
View Code

 

EOJ 2019.2月赛 D. 进制转换

原文:https://www.cnblogs.com/ckxkexing/p/10422291.html

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