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[1000005];

int solve(int tot){
    int i=1;
    while(i<=tot)
    {
        while(i<=tot&&p[i].num>p[i+1].num)
        {
            k-=p[i].sum;
            if(k<=0)
            {
                return p[i].firstX;    
            }
            p[i].sum=0;
            i++;    
        }
        i++;    
    }
    
    for(i=tot;i>=1;--i)
    {
        k-=p[i].sum;
        if(k<=0)
        {
            return p[i].firstX;    
        }
        p[i].sum=0;
    }
    return 0;

}


int main()
{
    p[0].num=0;
    while(~scanf("%d %d",&n,&k))
    {
        char c;scanf("%c",&c);
        int tot=0;
        for(int i=1;i<=n;++i)
        {
            scanf("%c",&c);
            if(p[tot].num==c-'0')
            {
                p[tot].sum++;
            }
            else 
            {
                tot++;
                p[tot].num=c-'0';
                p[tot].sum=1;
                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;
        
    
    }

}