1124 咸鱼魔法记(“玲珑杯”线上赛 Round #15 河南专场)

题目链接：http://ifrog.cc/acm/problem/1124

这道题两种思路

• 1.二分答案加O(n)判断
• 2.直接O(n)尺取法，尺取出最大答案

二分代码：

#include<bits/stdc++.h>
using namespace std;

int arr[1000007];
int n, k;

bool judge(int x)
{
    int res = 0;
    for(int i=0; i<x; ++ i)
        res += arr[i];
    if(res + k >= x)
        return true;
    for(int i=x; i<n; ++ i)
    {
        res += arr[i] - arr[i-x];
        if(res + k >= x)
            return true;
    }
    return false;
}

int main()
{
    while(scanf("%d%d", &n, &k) != EOF)
    {
        for(int i=0; i<n; ++ i)
            scanf("%d", &arr[i]);
        int b = 0, e = n;
        while(b < e)
        {
            int mid = b + (e - b + 1) / 2;
            if(judge(mid))
                b = mid;
            else
                e = mid - 1;
        }
        printf("%d\n", b);
    }
    return 0;
}

尺取代码如下：

#include<bits/stdc++.h>
using namespace std;


int arr[300007];
int main()
{
    int n, k;
    while(scanf("%d%d", &n, &k)!=EOF)
    {
        for(int i=1; i<=n; ++ i)
            scanf("%d", &arr[i]);

        int mx = 0;
        if(k == 0)
        {
            int res = 0;
            for(int i=1; i<=n; ++ i)
            {
                if(arr[i] == 1)
                    res ++;
                else
                {
                    mx = max(mx, res);
                    res = 0;
                }
            }
            printf("%d\n", res);
            continue;
        }

        for(int b = 1, e = 0, res = 0; b <= n;)
        {
            while(e+1 <= n && res + (arr[e + 1] == 0) <= k)
                res += (arr[++ e] == 0);

            mx = max(mx, e - b + 1);
            if(e + 1 > n)
                break;
            while(res == k && b <= e)
            {
                res -= (arr[b ++] == 0);
                if(res < k)
                    break;
            }
        }
        printf("%d\n", mx);
    }
    return 0;
}