Problem:
Score : 500 points

Problem Statement

We have N bulbs arranged on a number line, numbered 1 to N from left to right.
Bulb i is at coordinate i.

Each bulb has a non-negative integer parameter called intensity.
When there is a bulb of intensity d at coordinate x, the bulb illuminates the segment from coordinate x-d-0.5 to x+d+0.5.
Initially, the intensity of Bulb i is A_i. We will now do the following operation K times in a row:

For each integer i between 1 and N (inclusive), let B_i be the number of bulbs illuminating coordinate i. Then, change the intensity of each bulb i to B_i.

Find the intensity of each bulb after the K operations.

Constraints

1 \leq N \leq 2 \times 10^5

1 \leq K \leq 2 \times 10^5

0 \leq A_i \leq N

Input

Input is given from Standard Input in the following format:

N K
A_1 A_2 \ldots A_N

Output

Print the intensity A{'}_i of each bulb i after the K operations to Standard Output in the following format:

A{'}_1 A{'}_2 \ldots A{'}_N

Sample Input 1

5 1
1 0 0 1 0

Sample Output 1

1 2 2 1 2

Initially, only Bulb 1 illuminates coordinate 1, so the intensity of Bulb 1 becomes 1 after the operation.
Similarly, the bulbs initially illuminating coordinate 2 are Bulb 1 and 2, so the intensity of Bulb 2 becomes 2.

Sample Input 2

5 2
1 0 0 1 0

Sample Output 2

3 3 4 4 3

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

#include <iostream>
#include <string>
#include <algorithm>
#include <vector>
//#include <stack>
//#include <queue>
//#include <cstdio>
#include <cmath>
#include <iterator>
#include <map>
//#include <list>
#include <iomanip>

using namespace std;

#define IOS ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
#define prec std::cout << std::fixed << std::setprecision(15);
#define endl "\n"
#define MOD 1000000007
#define Int int64_t
#define PI 3.14159265358979
#define ssort(z) sort(z.begin(), z.end())
#define rsort(z) sort(z.rbegin(), z.rend())
#define eerase(z) z.erase(unique(z.begin(), z.end()), z.end())
#define ccnt(z, w) count(z.begin(), z.end(), w)
#define rep(i,a,n) for(Int (i)=(a); (i)<(n); (i)++)
#define repq(i,a,n) for(Int (i)=(a); (i)<=(n); (i)++)
const int MAX_N = 1000000;
const Int MAX_N_Int = 1000000000000;

template <typename T>
void printV(const std::vector<T>& v, const char * const separator = " ")
{
    if(!v.empty())
    {
        std::copy(v.begin(),
                  --v.end(),
                  std::ostream_iterator<T>(std::cout, separator));
        std::cout << v.back() << "\n";
    }
}

bool isPrime(int num)
{
    if (num < 2) return false;
    else if (num == 2) return true;
    else if (num % 2 == 0) return false; // 偶数はあらかじめ除く

    double sqrtNum = sqrt(num);
    for (int i = 3; i <= sqrtNum; i += 2)
    {
        if (num % i == 0)
        {
            // 素数ではない
            return false;
        }
    }

    // 素数である
    return true;
}
Int gcd(Int a, Int b) {
    return b != 0 ? gcd(b, a % b) : a;
}

Int lcm(Int a, Int b) {
    return a / gcd(a, b) * b;
    //a*bは64bit integer overflow
}

int Max(int a, int b, int c) {
    int temp = max(a, b);
    return max(temp, c);
}

int Min(int a, int b, int c) {
    int temp = min(a, b);
    return min(temp, c);
}

bool integer(double num) {
    return floor(num) == num;
}

Int fact(int num) {
    if (num == 0)
        return 1;
    else
        return num * fact(num - 1);
}

Int yakusu(int n) {
    int cnt = 0;
    for (int i = 1; i * i <= n; i++) {
        if (n % i == 0) {
            cnt++;
            if (i * i != n) cnt++;
        }
    }
    return cnt;
}


Int fact_mod(Int n, Int mod) {
    Int f = 1; repq(i,2,n) f = f * (i % MOD) % MOD;
    return f;
}

// 繰り返し二乗法 (modの世界での累乗)
Int mod_pow(Int x, Int n, Int mod) {
    Int res = 1;
    while(n > 0) {
        if(n & 1) res = (res * x) % mod;
        x = (x * x) % mod;
        n >>= 1;
    }
    return res;
}

// 組み合わせ nCr を求める
Int combination_mod(Int n, Int r, Int mod) {
    if(r > n-r) r = n-r;
    if(r == 0) return 1;
    Int a = 1;
    rep(i, 0, r) a = a * ((n-i) % mod) % mod;
    Int b = mod_pow(fact_mod(r, mod), mod-2, mod);
    return (a % mod) * (b % mod) % mod;
}

int main() {
    //COMinit();
    IOS;
    prec;
    Int a,b,c,n,m,k,x,y=0,ans=1;
    string str,s="",t,u;
    cin>>n>>k;
    vector<Int>A(n),zero(n),mem(n);
    for (int i = 0; i < n; ++i) {
        cin>>A[i];
    }
    vector<Int>imos(n+1),res(n);
    for (int o = 0;  o<k ; o++) {
        for (Int j = 0; j < n; ++j) {
            Int l=max((Int)0,j-A[j]), r=min(n-1,j+A[j]);
            //cout<<l<<" "<<r<<endl;
            imos[l]+=1;
            imos[r+1]-=1;
        }
        res[0]=imos[0];
        for (int i = 1; i < n; ++i) {
            res[i]+=res[i-1]+imos[i];
        }
        A=res;
        imos=zero;
        if(res==mem){
            break;
        }
        mem=res;
        if(o!=k-1)res=zero;
    }

    printV(res);





    cout<<endl;
    return 0;
}