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
Memory limit: 128 MB
In a cinema there is a long row of NN seats. Some of them are occupied, some of them are free. At most one of the people who are already seated can move to another seat.
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 NN. The 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
- 2 leq N leq 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 44th seat to the 77th, obtaining the configuration
|
001100
|
3
|
One way is to move the person from the 33rd seat to the 55th one. This way, the cinema row becomes
|
01000100
|
6
|
Move the person from the 66th seat to the 11st one. This way, it's posible to sear a group of size 66 |