zyb的面试

zyb的面试

http://acm.hdu.edu.cn/showproblem.php?pid=6468

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 470    Accepted Submission(s): 179

Problem Description

今天zyb参加一场面试,面试官听说zyb是ACMer之后立马抛出了一道算法题给zyb:
有一个序列,是1到n的一种排列,排列的顺序是字典序小的在前,那么第k个数字是什么?
例如n=15,k=7, 排列顺序为1, 10, 11, 12, 13, 14, 15, 2, 3, 4, 5, 6, 7, 8, 9;那么第7个数字就是15.
那么,如果你处在zyb的场景下,你能解决这个问题吗?

 

Input

T组样例(T<=100)
两个整数n和k(1<=n<=1e6,1<=k<=n),n和k代表的含义如上文

 

Output

输出1-n之中字典序第k小的数字

 

Sample Input

1

15 7

 

Sample Output

15

暴力模拟

#include<bits/stdc++.h>
using namespace std;
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define sqr(x) ((x)*(x))
#define pb push_back
#define eb emplace_back
#define maxn 100005
#define eps 1e-8
#define pi acos(-1.0)
#define rep(k,i,j) for(int k=i;k<j;k++)
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<double,double>pdd;
typedef pair<int,char> pic;
typedef pair<pair<int,string>,pii> ppp;
typedef unsigned long long ull;
const long long mod=1e9+7;
const double oula=0.57721566490153286060651209;
using namespace std;

int n,k;

int main(){

    int t;
    cin>>t;
    while(t--){
        cin>>n>>k;
        int ans=1;
        k--;
        while(k){
            if(ans*10<=n){
                ans*=10;
                k--;
            }
            else{
                if(ans+1<=n){
                    while(ans%10==9) ans/=10;
                    ans++;
                    k--;
                }
                else{
                    ans/=10;
                    while(ans%10==9) ans/=10;
                    ans++;
                    k--;
                }
            }
        }
        cout<<ans<<endl;
    }

}