[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。
1 #pragma comment(linker, "/STACK:10240000,10240000") 2 3 #include <iostream> 4 #include <cstdio> 5 #include <algorithm> 6 #include <cstdlib> 7 #include <cstring> 8 #include <map> 9 #include <queue> 10 #include <deque> 11 #include <cmath> 12 #include <vector> 13 #include <ctime> 14 #include <cctype> 15 #include <set> 16 #include <bitset> 17 #include <functional> 18 #include <numeric> 19 #include <stdexcept> 20 #include <utility> 21 22 using namespace std; 23 24 #define mem0(a) memset(a, 0, sizeof(a)) 25 #define lson l, m, rt << 1 26 #define rson m + 1, r, rt << 1 | 1 27 #define define_m int m = (l + r) >> 1 28 #define rep0(a, b) for (int a = 0; a < (b); a++) 29 #define rep1(a, b) for (int a = 1; a <= (b); a++) 30 #define all(a) (a).begin(), (a).end() 31 #define lowbit(x) ((x) & (-(x))) 32 #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) {} 33 #define constructInt3(name, a, b, c) name(int a = 0, int b = 0, int c = 0): a(a), b(b), c(c) {} 34 #define constructInt2(name, a, b) name(int a = 0, int b = 0): a(a), b(b) {} 35 #define pchr(a) putchar(a) 36 #define pstr(a) printf("%s", a) 37 #define sint(a) ReadInt(a) 38 #define sint2(a, b) ReadInt(a);ReadInt(b) 39 #define sint3(a, b, c) ReadInt(a);ReadInt(b);ReadInt(c) 40 #define pint(a) WriteInt(a) 41 42 typedef double db; 43 typedef long long LL; 44 typedef pair<int, int> pii; 45 typedef multiset<int> msi; 46 typedef set<int> si; 47 typedef vector<int> vi; 48 typedef map<int, int> mii; 49 50 const int dx[8] = {0, 1, 0, -1, 1, 1, -1, -1}; 51 const int dy[8] = {1, 0, -1, 0, -1, 1, 1, -1}; 52 const int maxn = 1e3 + 7; 53 const int maxm = 1e5 + 7; 54 const int maxv = 1e7 + 7; 55 const int max_val = 1e6 + 7; 56 const int MD = 1e9 +7; 57 const int INF = 1e9 + 7; 58 const double PI = acos(-1.0); 59 const double eps = 1e-10; 60 61 template<class T>T gcd(T a, T b){return b==0?a:gcd(b,a%b);} 62 template<class T>void ReadInt(T &x){char c=getchar();while(!isdigit(c))c=getchar();x=0;while(isdigit(c)){x=x*10+c-'0';c=getchar();}} 63 template<class T>void WriteInt(T i) {int p=0;static int b[20];if(i == 0) b[p++] = 0;else while(i){b[p++]=i%10;i/=10;}for(int j=p-1;j>=0;j--)pchr('0'+b[j]);} 64 65 66 LL work(int a, int b) { 67 if (b <= a) return 0; 68 return max(0, (a + b - 1) / 2 - (b - a) / 2); 69 } 70 LL work(int maxv) { 71 LL ans = 0; 72 rep1(i, maxv - 1) { 73 ans += work(i, maxv - i); 74 } 75 return ans; 76 } 77 int a[1010]; 78 int main() { 79 //freopen("in.txt", "r", stdin); 80 int n, m; 81 while (cin >> n >> m) { 82 rep0(i, m) { 83 sint(a[i]); 84 } 85 sort(a, a + m); 86 int minv = a[0] - 1, maxv = n - a[m - 1]; 87 if (minv > maxv) swap(minv, maxv); 88 if (minv == 0) pint(work(maxv)); 89 else { 90 if (minv == 1) { 91 if (maxv & 1) pint(0); 92 else pint(1); 93 } 94 else { 95 if (minv == maxv) pint(0); 96 else pint(work(minv, maxv)); 97 } 98 } 99 pchr('\n'); 100 } 101 return 0; 102 }
