思路：把一个木棍分成3段，使之能够构成三角形的方案总数可以这样计算，枚举一条边，然后可以推公式算出当前方案数。对于已知一条边的情况，也用公式推出。用max和min并维护下，以减少情况数目。
#pragma comment(linker, "/STACK:10240000,10240000") #include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstdlib>
#include <cstring>
#include <map>
#include <queue>
#include <deque>
#include <cmath>
#include <vector>
#include <ctime>
#include <cctype>
#include <set>
#include <bitset>
#include <functional>
#include <numeric>
#include <stdexcept>
#include <utility> using namespace std; #define mem0(a) memset(a, 0, sizeof(a))
#define lson l, m, rt << 1
#define rson m + 1, r, rt << 1 | 1
#define define_m int m = (l + r) >> 1
#define rep0(a, b) for (int a = 0; a < (b); a++)
#define rep1(a, b) for (int a = 1; a <= (b); a++)
#define all(a) (a).begin(), (a).end()
#define lowbit(x) ((x) & (-(x)))
#define constructInt4(name, a, b, c, d) name(int a = 0, int b = 0, int c = 0, int d = 0): a(a), b(b), c(c), d(d) {}
#define constructInt3(name, a, b, c) name(int a = 0, int b = 0, int c = 0): a(a), b(b), c(c) {}
#define constructInt2(name, a, b) name(int a = 0, int b = 0): a(a), b(b) {}
#define pchr(a) putchar(a)
#define pstr(a) printf("%s", a)
#define sint(a) ReadInt(a)
#define sint2(a, b) ReadInt(a);ReadInt(b)
#define sint3(a, b, c) ReadInt(a);ReadInt(b);ReadInt(c)
#define pint(a) WriteInt(a) typedef double db;
typedef long long LL;
typedef pair<int, int> pii;
typedef multiset<int> msi;
typedef set<int> si;
typedef vector<int> vi;
typedef map<int, int> mii; const int dx[] = {, , , -, , , -, -};
const int dy[] = {, , -, , -, , , -};
const int maxn = 1e3 + ;
const int maxm = 1e5 + ;
const int maxv = 1e7 + ;
const int max_val = 1e6 + ;
const int MD = 1e9 +;
const int INF = 1e9 + ;
const double PI = acos(-1.0);
const double eps = 1e-; template<class T>T gcd(T a, T b){return b==?a:gcd(b,a%b);}
template<class T>void ReadInt(T &x){char c=getchar();while(!isdigit(c))c=getchar();x=;while(isdigit(c)){x=x*+c-'';c=getchar();}}
template<class T>void WriteInt(T i) {int p=;static int b[];if(i == ) b[p++] = ;else while(i){b[p++]=i%;i/=;}for(int j=p-;j>=;j--)pchr(''+b[j]);} LL work(int a, int b) {
if (b <= a) return ;
return max(, (a + b - ) / - (b - a) / );
}
LL work(int maxv) {
LL ans = ;
rep1(i, maxv - ) {
ans += work(i, maxv - i);
}
return ans;
}
int a[];
int main() {
//freopen("in.txt", "r", stdin);
int n, m;
while (cin >> n >> m) {
rep0(i, m) {
sint(a[i]);
}
sort(a, a + m);
int minv = a[] - , maxv = n - a[m - ];
if (minv > maxv) swap(minv, maxv);
if (minv == ) pint(work(maxv));
else {
if (minv == ) {
if (maxv & ) pint();
else pint();
}
else {
if (minv == maxv) pint();
else pint(work(minv, maxv));
}
}
pchr('\n');
}
return ;
}