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; } }
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