C.Necklace

C. Necklace
time limit per test1.0 s
memory limit per test512 megabytes
inputstandard input
outputstandard output
Bob has given Alice a necklace as her birthday gift. That necklace has N crystals, M of which catches her fancy. Formally, crystals with labels of 1…N are strung sequentially in a circle. And those catching her fancy are labelled n1,n2,…,nM.

Alice would like the necklace to be divided into M pieces. She asked Bob to sever the necklace into M pieces. Each piece should consist of a sequence of crystals as in the original necklace (i.e., the crystal labels are contiguous, except that 1 following N is accepted) and should include one crystal catching Alice’s fancy. Obviously, each crystal must belong to exactly one piece.

However, an excessively long piece of severed necklaces could not fit Alice’s slim neck. So she wants to know, among all possible severings as requested, what the minimum length of the longest piece from a severed necklace is. She wants you to answer this question.

Input
The first line of the input data contains 2 integers N and M (1≤M≤N<1018 and M≤106).

The second line contains M integers, the i-th of which represents ni. The ni’s are given in ascending order.

Output
You need to output your answer in one line.

Example
inputCopy
10 4
2 5 6 8
outputCopy
3
Note
As for the example: You can sever the necklace into [1,3],[4,5],[6,7],[8,10].

Considering huge scale of data, here is a way provided to help input faster:

#define gc()(is==it?it=(is=in)+fread(in,1,Q,stdin),(is==it?EOF:*is++):*is++)
const int Q=(1«24)+1;
char in[Q],*is=in,*it=in,c;
void read(long long &n){for(n=0;(c=gc())<'0'||c>'9';);for(;c<='9'&&c>='0';c=gc())n=n*10+c-48;
}

题意：

• 长度为 n 的项链（是个环！）上有 m 个宝石，每个宝石要放在同一段里，求最长段长度的最小值。

思路：

• 开始没看到是环，一眼二分答案想硬A上去，WA了两发才发现还有环这个操作。不过肯定还是二分答案+贪心的，就是贪心判断的方法需要改一下。

• 每次check的时候从第一个位置开始贪心，一开始最大偏移量n-a[m]+a[1]-1，开个变量表示当前起始位置最大的偏移量，每跑一段区间都取个min，然后记录一下最开始的起始位置一共最大往前的偏移量即可。

• 官方讲解：
  
  

  #include
  using namespace std;
  typedef long long LL;
  #define gc()(is==it?it=(is=in)+fread(in,1,Q,stdin),(is==it?EOF:*is++):*is++)
  const int Q=(1<<24)+1;
  char in[Q],*is=in,*it=in,c;
  void read(long long &n){for(n=0;(c=gc())<'0'||c>'9';);for(;c<='9'&&c>='0';c=gc())n=n*10+c-48;
  }const int maxn = 1e6+10;
  LL n, m, a[maxn];
  int check(LL x){LL mx = n-(a[m]-a[1]+1), mm = 0;   //最大能往左边移的量, 当前左移的量LL lst = 0, mx0 = mx;              //后一段要给前一段的大小for(int i = 1; i < m; i++){LL now = x-lst;           //当前能分的=x-给了上一段的if(now<=0)return 0;       //当前能分的是负数或0了，寄LL cc = a[i+1]-a[i];      //当前需要分的if(cc>now){               //足够分lst = cc-now;         //分一下mx = min(mx, now-1);  //更新最大偏移}else{lst = 0;                 //不够分LL m3 = min(mx, now-cc); //往左边移mm += m3;                //累加左移的量mx = min(mx-m3, cc-1);   //更新最大偏移}}LL ed = lst+mx0+1-x;if(x-lst>=mx0+1 || mm>=ed)return 1;return 0;
  }int main(){read(n);  read(m);for(int i = 1; i <= m; i++)read(a[i]);LL l = 1, r = n; while(l < r){LL mid = (l+r)>>1;if(check(mid))r = mid;else l = mid+1;}cout<

#include
using namespace std;
typedef long long LL;
const int maxn = 3e5+10;
const LL mod = 998244353;
LL f[maxn];
int main(){LL n;  cin>>n;f[1] = f[2] = 1;for(int i = 3; i <= n; i++){f[i] = (f[i-1]+f[i-2])%mod;}cout<