bzoj4029[HEOI2015]定价

http://www.lydsy.com/JudgeOnline/problem.php?id=4029

贪心。

枚举有多少个后导0（不妨枚举到有k个后导0），找到第一个大于等于L的$10^k$的倍数，和第一个大于等于L的$5*10^k$的奇数次倍数，只有这2个才有可能成为答案。

然后从中选出最优即可。

#include<cstdio>
#include<cstdlib>
#include<iostream>
#include<fstream>
#include<algorithm>
#include<cstring>
#include<string>
#include<cmath>
#include<queue>
#include<stack>
#include<map>
#include<utility>
#include<set>
#include<bitset>
#include<vector>
#include<functional>
#include<deque>
#include<cctype>
#include<climits>
#include<complex>
//#include<bits/stdc++.h>适用于CF,UOJ，但不适用于poj
 
using namespace std;

typedef long long LL;
typedef double DB;
typedef pair<int,int> PII;
typedef complex<DB> CP;

#define mmst(a,v) memset(a,v,sizeof(a))
#define mmcy(a,b) memcpy(a,b,sizeof(a))
#define fill(a,l,r,v) fill(a+l,a+r+1,v)
#define re(i,a,b)  for(i=(a);i<=(b);i++)
#define red(i,a,b) for(i=(a);i>=(b);i--)
#define ire(i,x) for(typedef(x.begin()) i=x.begin();i!=x.end();i++)
#define fi first
#define se second
#define m_p(a,b) make_pair(a,b)
#define p_b(a) push_back(a)
#define SF scanf
#define PF printf
#define two(k) (1<<(k))

template<class T>inline T sqr(T x){return x*x;}
template<class T>inline void upmin(T &t,T tmp){if(t>tmp)t=tmp;}
template<class T>inline void upmax(T &t,T tmp){if(t<tmp)t=tmp;}

const DB EPS=1e-9;
inline int sgn(DB x){if(abs(x)<EPS)return 0;return(x>0)?1:-1;}
const DB Pi=acos(-1.0);

inline int gint()
  {
        int res=0;bool neg=0;char z;
        for(z=getchar();z!=EOF && z!='-' && !isdigit(z);z=getchar());
        if(z==EOF)return 0;
        if(z=='-'){neg=1;z=getchar();}
        for(;z!=EOF && isdigit(z);res=res*10+z-'0',z=getchar());
        return (neg)?-res:res; 
    }
inline LL gll()
  {
      LL res=0;bool neg=0;char z;
        for(z=getchar();z!=EOF && z!='-' && !isdigit(z);z=getchar());
        if(z==EOF)return 0;
        if(z=='-'){neg=1;z=getchar();}
        for(;z!=EOF && isdigit(z);res=res*10+z-'0',z=getchar());
        return (neg)?-res:res; 
    }

const int INF=0x3f3f3f3f;

LL A,B,ans;
int x;

inline int col(LL v)
  {
      int res=0;
      while(v%10==0)v/=10;
      if(v%10==5)res--;
      while(v!=0)v/=10,res+=2;
      return res;
  }

inline void DFS(LL div)
  {
      int C=((A-1)/div+1)*div,dc=col(C);
      int D=(((A-1)/(5*div)+1)&1) ? ((A-1)/(5*div)+1)*(5*div) : ((A-1)/(5*div)+1+1)*(5*div),dd=col(D);
      if(C>B)return;
      if(D>B)dd=INF;
      if(dc<x)x=dc,ans=C;
      if(dd<x)x=dd,ans=D;
      DFS(div*10);
  }

int main()
  {
      freopen("bzoj4029.in","r",stdin);
      freopen("bzoj4029.out","w",stdout);
      for(int Case=gint();Case;Case--)
        {
            cin>>A>>B;
            x=INF;
            DFS(1);
            cout<<ans<<endl;
        }
      return 0;
  }