[2016-04-09][codeforces][660][C][Hard Process]

• 时间:2016-04-09 23:59:40 星期六

• 题目编号:[2016-04-09][codeforces][660][C][Hard Process]

• 题目大意:给定一个0 1序列,在最多把k个0改成1的情况下,最长的1子串有多长,

• 分析:

  • 二分答案
    • check():枚举长度为mid的子串,判断里面的0的数目是否小于等于1,

#include<cstdio>
using namespace std;
const int maxn = 3*1E5 + 10;
int a[maxn],b[maxn],n,k;
int check(int x){
        for(int i = x ; i <= n ; ++i){
                if(b[i] - b[i - x] + k >= x)    return 1;
        }
        return 0;
}
void print(int x){
        int flg = 0;
        for(int i = x ; i <= n ; ++i){
                if(b[i] - b[i - x] + k >= x){
                        for(int j = i - x + 1;j <= i ; ++j){
                                a[j] = 1;flg = 1;
                        }
                }
                if(flg)   break;
        } 
        for(int i = 1;i <= n ; ++i)       printf("%d ",a[i]);
}
int main(){
        scanf("%d%d",&n,&k);
        for(int i = 1 ; i <= n ; ++i){
                scanf("%d",a+i);
                b[i] = b[i - 1] + a[i];
        }
        int l = 0,r = n + 1,mid,res;
        while(r >= l){
                mid = (l + r)>>1;
                if(check(mid)){
                        l = mid + 1;
                        res = mid;
                }
                else r = mid - 1;               
        }
        printf("%d\n",res);
        print(res);
        return 0;
}

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