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UVa 11401 Triangle Counting (计数DP)
题意:给定一个数 n,从1-n这些数中任意挑出3个数,能组成三角形的数目。
析:dp[i] 表示从1-i 个中任意挑出3个数,能组成三角形的数目。
代码如下:
#pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostream> #include <cstring> #include <set> #include <queue> #include <algorithm> #include <vector> #include <map> #include <cctype> #include <cmath> #include <stack> #define freopenr freopen("in.txt", "r", stdin) #define freopenw freopen("out.txt", "w", stdout) using namespace std; typedef long long LL; typedef unsigned long long ULL; typedef pair<int, int> P; const int INF = 0x3f3f3f3f; const double inf = 0x3f3f3f3f3f3f; const LL LNF = 0x3f3f3f3f3f3f; const double PI = acos(-1.0); const double eps = 1e-8; const int maxn = 1e6 + 5; const int mod = 1e9 + 7; const int dr[] = {-1, 0, 1, 0}; const int dc[] = {0, 1, 0, -1}; const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"}; int n, m; const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; inline int Min(int a, int b){ return a < b ? a : b; } inline int Max(int a, int b){ return a > b ? a : b; } inline LL Min(LL a, LL b){ return a < b ? a : b; } inline LL Max(LL a, LL b){ return a > b ? a : b; } inline int gcd(int a, int b){ return b ? gcd(b, a%b) : a; } inline int lcm(int a, int b){ return a * b / gcd(a, b); } inline bool is_in(int r, int c){ return r >= 0 && r < n && c >= 0 && c < m; } LL dp[maxn]; int main(){ dp[3] = 0; for(int i = 4; i < maxn; ++i){ if(i & 1) dp[i] = dp[i-1] + (LL)(i-1)*((i-4)/2+1)/2; else dp[i] = dp[i-1] + (LL)((i-3)/2+1)*(i-2)/2; } while(cin >> n && n > 2){ cout << dp[n] << endl; } return 0; }
UVa 11401 Triangle Counting (计数DP)
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