codeforces 1177B

题目链接：http://codeforces.com/problemset/problem/1177/B

B. Digits Sequence (Hard Edition)

Let's write all the positive integer numbers one after another from 11 without any delimiters (i.e. as a single string). It will be the infinite sequence starting with 123456789101112131415161718192021222324252627282930313233343536...

Your task is to print the kk-th digit of this sequence.

Input

The first and only line contains integer kk (1≤k≤10121≤k≤1012) — the position to process (11-based index).

Output

Print the kk-th digit of the resulting infinite sequence.

Examples

input

output

input

output

题意概括：把正整数按顺序排成一个字符串，求第n个字符是什么

解题思路：一位正整数有9个，两位有90个，三位900个，先求出这个数是几位的，然后求出是第几个，最后求出是这个数的第几位

#include<iostream>
#include<queue>
#include<cstring>
#include<cmath>
#include<map>
#include<algorithm>
#define up(i,x,y) for(i=x;i<=y;i++)
#define down(i,x,y) for(i=x;i>=y;i--)
#define MAX(a,b) a>b?a:b
#define MIN(a,b) a<b?a:b
#define MAX(a,b,c) (a>b?(a>c?a:c):(b>c?b:c))
#define MIN(a,b,c) (a<b?(a<c?a:c):(b<c?b:c))
using namespace std;typedef long long ll;
const double PI = acos(-1.0);
const double eps = 1e-6;
const int INF = 1000000000;
const int maxn = 100;
const int MAXN = 2750131;
ll T,n,m,i,j;
ll gcd(ll p,ll q)
{return q==0?p:gcd(q,p%q);}ll g[200005];int main()
{ios::sync_with_stdio(false);ll a[20];ll k=1,s=1,sum=9;for(i=0;s<1e12;i++){a[i]=s*sum*k;s*=10;k++;//   cout<<a[i]<<endl;}cin>>n;if(n<10){cout<<n<<endl;return 0;}sum=0;i=0;k=1;s=1;ll m=n;while(m>a[i]){m-=a[i];i++;k++;s=pow(10,i);sum=s-1;}
//  cout<<m<<endl;ll ans=(m)/k+sum;if(m%k){ans++;}
//  cout<<ans<<endl;ll sk=m%k;if(sk==0){sk=k;}//cout<<sk<<" "<<k<<endl;while(ans>0){// cout<<ans<<" "<<k<<" "<<sk<<endl;if(sk==k){cout<<ans%10<<endl;break;}else{ans/=10;k--;}} return 0;
}

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