CS Academy Round41 Cinema Seats

题目链接：https://csacademy.com/contest/archive/task/cinema-seats

题目大意：给你一串长度为n的01字符串表示n个座位，0表示有人，1表示没人，移动一个人表示将1换到另一个为0的地方。问在只允许移动一个人或0个人情况下连续座位的最大长度。

解题思路：考虑00100101的前缀和00111223，假设移动了第三个1，那么前缀和变为00000234，即是说我们只要找到前缀和为aaa(a+1)(a+1)(a+1)的最大长度即可。那么就可以枚举然后二分来做了。

需要注意的是当空位的个数比找到的长度小的时候要将结果-1，实际上是这个1被移动到了找到的长度的最末尾。

代码：

 1 const int maxn = 1e5 + 5;
 2 char str[maxn];
 3 int a[maxn];
 4 
 5 int main(){
 6     scanf("%s", str);
 7     int len = strlen(str), zeronum = 0;
 8     a[0] = (str[0] == '0'? 0: 1);
 9     zeronum += (str[0] == '0'? 1: 0);
10     for(int i = 1; i < len; i++){
11         a[i] = a[i - 1] + (str[i] == '0'? 0: 1);
12         zeronum += (str[i] == '0'? 1: 0);
13     }
14     int maxv = 0, maxlen = 0, m = a[len - 1];
15     
16     for(int i = a[0]; i <= m; i++){
17         int tmlen = lower_bound(a, a + len, i + 2) - lower_bound(a, a + len, i);
18         if(tmlen > maxlen){
19             maxlen = tmlen;
20             maxv = i;
21         }
22     }
23     if(maxv != 0) maxlen--;
24     if(zeronum < maxlen) maxlen--;
25     printf("%d
", maxlen); 
26 }

题目：

Cinema Seats

Time limit: 1000 ms
Memory limit: 128 MB

In a cinema there is a long row of

A group of friends are going to come the cinema. What is the largest size of the group such that they can take a contiguous sequence of seats?

Standard input

The first line contains a binary string of length $i^{th}ith character is 0 is the i^{th}ith seat is empty, and 1 if the seat is taken.$

Standard output

Print the answer on the first line.

Constraints and notes

$10^52≤N≤105$
There will be at least one taken seat
There will be at least one free seat

Input	Output	Explanation
10010101	4	You can move the person seating in the `12222111` - the group's chairs were marked with `2`
001100	3	One way is to move the person from the This way, the cinema row becomes `000110`, leaving a sequence of
01000100	6	Move the person from the This way, it's posible to sear a group of size