Codeforces 1204D2 Kirk and a Binary String (hard version)

D1. Kirk and a Binary String (easy version)

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

The only difference between easy and hard versions is the length of the string. You can hack this problem only if you solve both problems.

Kirk has a binary string ss (a string which consists of zeroes and ones) of length nn and he is asking you to find a binary string tt of the same length which satisfies the following conditions:

• For any ll and rr (1≤l≤r≤n1≤l≤r≤n) the length of the longest non-decreasing subsequence of the substring slsl+1…srslsl+1…sr is equal to the length of the longest non-decreasing subsequence of the substring tltl+1…trtltl+1…tr;
• The number of zeroes in tt is the maximum possible.

A non-decreasing subsequence of a string pp is a sequence of indices i1,i2,…,iki1,i2,…,ik such that i1<i2<…<iki1<i2<…<ik and pi1≤pi2≤…≤pikpi1≤pi2≤…≤pik. The length of the subsequence is kk.

If there are multiple substrings which satisfy the conditions, output any.

Input

The first line contains a binary string of length not more than 20002000.

Output

Output a binary string which satisfied the above conditions. If there are many such strings, output any of them.

题意
给出一个 串 S ,求一个串 T , 要求LIS等长，所有区间的 LIS  相等，0 的个数尽可能多

解:

从后往前贪心， 保证后面的解都是合法的

 假如当前是0 作为起点 对后面的Lis 无影响

假如当前是 1   假如不是以他为起点 对于LIS 起点为0 的 那么 变为0 就一定更改lis  但是1的个数>= 0 的个数时可以改

假如当前是 1   假如是以他为起点 那么 就变为0 就一定不会更改lis  

所以 但是当 $ tot1>=tot0  $ 时 可以没有影响 可以改

//
#include<bits/stdc++.h>
using namespace std;
string s;
int main()
{
    cin>>s;
    int tot1=0,tot2=0;
    for(int i=s.size()-1;i>=0;i--)
    {
        if(s[i]=='1') if(tot1>=tot2) s[i]='0';else tot1++;
        else
        {
            tot2++;
        }
        
    }
    cout<<s;
    
}