Codeforces 660C-Hard Process【尺取法练习】
C. Hard Process time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output You are given an array a with n 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 numbers one, as f(a). You can change no more than k zeroes to ones to maximize f(a).
Input The first line contains two integers n 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 ≤ ai ≤ 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 a after the changes.
If there are multiple answers, you can print any one of them.
题意:就是给你一串01字符串,然后你可以将其中k个0改成1,让你求最长的由1组成的字符串!
思路:有关区间动态变化的,很容易想到尺取法,但是我还是一直W,说明理解不够深,应用不行,练习太少!!!
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int num[300000+100];
int main()
{
int n,k;
while(scanf("%d%d",&n,&k)!=EOF)
{
for(int i=1;i<=n;i++)
{
scanf("%d",&num[i]);
}
int r=1,l=1,zr=0,ans=0,ansl=1,ansr=0;
while(r<=n)
{
while(r<=n&&zr<=k)
{
if(num[r]==0)
{
if(zr==k)
break;
else
{
zr++;
}
}
r++;
}
if(r-l>ans)
{
ansr=r-1;
ansl=l;
ans=r-l;
}
while(l<=n&&num[l])
l++;
l++,zr--;
}
printf("%dn",ans);
for(int i=1;i<=n;i++)
{
if(i>=ansl&&i<=ansr)
printf("1 ");
else
printf("%d ",num[i]);
}
printf("n");
}
return 0;
}
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