Hard Process (two points)

Source: Internet
Author: User
Tags cmath

Hard ProcessTime limit:1000MS Memory Limit:262144KB 64bit IO Format:%i64d & Amp %i64u Submit Status Practice codeforces 660C

Description

You is given an array  a  with  n Span class= "Apple-converted-space" > elements. Each element Of  a  is either 0 or 1.

Let's denote the length of the longest subsegment of consecutive elements in a, consisting of only number s one, as F(a). You can change no more than K zeroes to ones to maximize F(a).

Input

The first line contains Integers  n Span class= "Apple-converted-space" > and  k    (1≤ n ≤3 105, 0≤ k n )-the number of elements In  a  and the Parameter  K .

The second line contains n integers ai (0≤ a i ≤ 1)-the elements of a.

Output

On the first line print a non-negative integer  Z  -the maximal value Of  F ( a )  after no more Than  K   Changes of zeroes to ones.

On the second line print n integers aJ -the elements of the array< c10> A after the changes.

If There is multiple answers, you can print any one of the them.

Sample Input

Input
7 1 1 0 0 1 1 0 1
Output
4 1 0 0 1 1 1 1
Input
10 2 1 0 0 1 0 1 0 1 0 1
Output
5 1 0 0 1 1 1 1 1 0 1
The main point: let the number of K change, so that the number of consecutive 1 is the largest, we can ask for 0 prefix and, and then through two points to find the location; I've been using violence for a long time ...
Two points:
#include <iostream>#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>using namespacestd;Const intINF =0x3f3f3f3f;#defineMem (x, y) memset (x,y,sizeof (x))Const intMAXN =300010;intNUM[MAXN];intA[MAXN];intL, N, K;intjsintx) {     for(inti =0; i + x <= n; i++){        if(Num[i + x]-num[i] <=k) {L= i +1; return true; }    }    return false;}intErfen (intLintR) {    intMid, ans;  while(L <=R) {Mid= (L + r) >>1; if(JS (mid)) {ans=mid; L= Mid +1; }        ElseR= Mid-1; }    returnans;}intMain () { while(~SCANF ("%d%d", &n, &k)) {        inttemp; memset (num,0,sizeof(num));  for(inti =1; I <= N; i++) {scanf ("%d", A +i); Num[i]= Num[i-1] + (a[i] = =0); }        intAns = Erfen (0, N);  for(inti = L; I < L + ans; i++) {A[i]=1; } printf ("%d\n", ans);  for(inti =1; I <= N; i++){            if(I! =1) printf (" "); printf ("%d", A[i]); }puts (""); }    return 0;}

Violence timeout:

#include <iostream>#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>using namespacestd;Const intINF =0x3f3f3f3f;#defineMem (x, y) memset (x,y,sizeof (x))Const intMAXN =1000100;intNUM[MAXN];intPOS[MAXN];intP[MAXN];intMain () {intN, K;  while(~SCANF ("%d%d", &n, &k)) {        intGG =0;  for(inti =0; I < n; i++) {scanf ("%d", Num +i); if(Num[i] = =1) GG =1; }        if(!GG) {printf ("%d\n", K);  for(inti =0; I < K; i++){                if(i) printf (" "); printf ("1"); }             for(inti = k; I < n; i++){                if(i) printf (" "); printf ("0"); }puts (""); Continue; }        intkg =0, ans =0, temp =0, cnt =0, TP =0;  for(inti =0; I < n; i++){            if(kg = =0&& Num[i] = =1) {Temp=0; KG=1; CNT=0;  for(intj = i; J < N; J + +){                                                if(Num[j] = =1) {Temp++; }                        Else{                            if(CNT +1> k) Break; Pos[cnt++] =J; Temp++; }                    }                        intj =i;  while(CNT < K && J >0) {pos[cnt++] = --J; Temp++; }                                            if(Ans <temp) {ans=temp; TP=CNT;  for(intj =0; J < TP; J + +) {P[j]=Pos[j]; }                    }                }            if(Num[i] = =0) kg =0; }         for(inti =0; i < TP; i++){        //printf ("%d", P[i]);Num[p[i]] =1; }//puts ("");printf"%d\n", ans);  for(inti =0; I < n; i++){            if(i) printf (" "); printf ("%d", Num[i]); } puts (""); }    return 0;}

Hard Process (two points)

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