武大OJ 574. K-th smallest

Description

Give you a number S of length n,you can choose a position and remove the number on it.After that,you will get a new number.

More formally,you choose a number x(1<=x<=n),then you will get the number Rx=S1S2…..Sx-1 Sx+1……Sn..The problem is what number x you choose will get k-th smallest Rx of all R.

If there are more than one answer,choose smallest x.

Input

First line of each case contains two numbers n and k.(2 ≤ k≤  n ≤ 1 000 000).

Next line contains a number of length n. Each position corresponds to a number of 1-9.

Output

Output x on a single line for each case.

Sample Input

10 5
6228814462
10 4
9282777691

Sample Output

10
5

考虑123456543212345这种数字，就可以找到规律了

 #include<iostream>
 #include<stdio.h>
 #include<math.h>
 #include<stdlib.h>
 #include<string.h>
 #include<algorithm>
 using namespace std;

 int n,k;
 typedef struct{
     int num,sum,firstX;
 }Point;
 Point p[];

 int solve(int tot){
     int i=;
     while(i<=tot)
     {
         while(i<=tot&&p[i].num>p[i+].num)
         {
             k-=p[i].sum;
             if(k<=)
             {
                 return p[i].firstX;
             }
             p[i].sum=;
             i++;
         }
         i++;
     }

     for(i=tot;i>=;--i)
     {
         k-=p[i].sum;
         if(k<=)
         {
             return p[i].firstX;
         }
         p[i].sum=;
     }
     return ;

 }

 int main()
 {
     p[].num=;
     while(~scanf("%d %d",&n,&k))
     {
         char c;scanf("%c",&c);
         int tot=;
         for(int i=;i<=n;++i)
         {
             scanf("%c",&c);
             if(p[tot].num==c-'')
             {
                 p[tot].sum++;
             }
             else
             {
                 tot++;
                 p[tot].num=c-'';
                 p[tot].sum=;
                 p[tot].firstX=i;
             }
         }

     //    for(int i=1;i<=tot;++i)
     //    cout<<p[i].num<<" "<<p[i].sum<<" "<<p[i].firstX<<endl;

         cout<<solve(tot)<<endl;

     }

 }