T3 syx with his evil sequences (easy version)

This title belongs to the topic just do it. . .

CONSIDERING greedy

Number (= x) in the current point is to be deleted when considering its nearest point how many steps required, the nearest you can delete the point 2 ^k , and this point is the point closest to the front of it, and it is in front of deleted point out there is a cur, so each number required (the current position minus 2 ^k -cur) times to reach, so come to the program:

#include<bits/stdc++.h>
using namespace std;
inline int read() {
    int x(0),neg(1);char ch(getchar());
    while(!isdigit(ch)) {
        if (ch=='-') neg=-1;
        ch=getchar();
    }
    while(isdigit(ch)) {
        x=(x<<1)+(x<<3)+(ch^48);
        ch=getchar();
    }
    return x*neg;
}
int n,k,x,cnt,cur;
int main() {
    int ans=0;
    n=read(),k=read();
    for (int i=1;i<=n;++i) {
        x=read();
        if (x==k) {
            int t=i-(1<<((int)log2(i)));
            if (t>cur) {
                ans+=t-cur;
                cur=t;
            }
            cut ++; 
            Why ++ ; 
        } 
    } 
    Printf ( " % d \ n " , ans + ent);
    return  0 ; 
}