[hdu5203]计数水题

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 ;
 }