250pt：
题意：在二维平面上，给定3种，左转、右转，以及前进一步，的指令序列循环执行，问整个移动过程是否是发散的。
思路：直接模拟一个周期，然后判断1.方向是否和初始时不同 2.是否在原来的点
满足其一便是收敛。具体画个图便明白
code:
1 #line 7 "SequenceOfCommands.cpp" 2 #include <cstdlib> 3 #include <cctype> 4 #include <cstring> 5 #include <cstdio> 6 #include <cmath> 7 #include <algorithm> 8 #include <vector> 9 #include <string> 10 #include <iostream> 11 #include <sstream> 12 #include <map> 13 #include <set> 14 #include <queue> 15 #include <stack> 16 #include <fstream> 17 #include <numeric> 18 #include <iomanip> 19 #include <bitset> 20 #include <list> 21 #include <stdexcept> 22 #include <functional> 23 #include <utility> 24 #include <ctime> 25 using namespace std; 26 27 #define PB push_back 28 #define MP make_pair 29 30 #define REP(i,n) for(i=0;i<(n);++i) 31 #define FOR(i,l,h) for(i=(l);i<=(h);++i) 32 #define FORD(i,h,l) for(i=(h);i>=(l);--i) 33 34 typedef vector<int> VI; 35 typedef vector<string> VS; 36 typedef vector<double> VD; 37 typedef long long LL; 38 typedef pair<int,int> PII; 39 const int g[4][2] = {{1, 0},{0,-1},{-1, 0},{0, 1}}; 40 41 class SequenceOfCommands 42 { 43 public: 44 string whatHappens(vector <string> commands) 45 { 46 string S = accumulate(commands.begin(), commands.end(), string("")); 47 int x = 0, y = 0, dir = 0; 48 int n = S.size(); 49 for (int i = 0; i < n; ++i){ 50 if (S[i] == ‘L‘) dir = (dir + 3) % 4; 51 if (S[i] == ‘R‘) dir = (dir + 1) % 4; 52 if (S[i] == ‘S‘){ 53 x += g[dir][0]; 54 y += g[dir][1]; 55 } 56 } 57 if (x == 0 && y == 0) return "bounded"; 58 if (dir != 0) return "bounded"; 59 return "unbounded"; 60 } 61 };
500pt:
题意:将圆周 n<= 10^6等分,编号0~n-1，现在上面选取m <= 10^5个点，给定整数a b c，选择的第i个点的下标为(a*a*i+b*i+c)%n,如果已经被选过则选下一个木有被选的点。问这些点能构成多少个直角三角形。
思路：直角三角形必须经过圆心。所以n为奇数必然无解。
否则先求出这m个点。每次求完后标记下，并用并查集加速。
code:
1 #line 7 "RightTriangle.cpp" 2 #include <cstdlib> 3 #include <cctype> 4 #include <cstring> 5 #include <cstdio> 6 #include <cmath> 7 #include <algorithm> 8 #include <vector> 9 #include <string> 10 #include <iostream> 11 #include <sstream> 12 #include <map> 13 #include <set> 14 #include <queue> 15 #include <stack> 16 #include <fstream> 17 #include <numeric> 18 #include <iomanip> 19 #include <bitset> 20 #include <list> 21 #include <stdexcept> 22 #include <functional> 23 #include <utility> 24 #include <ctime> 25 using namespace std; 26 27 #define PB push_back 28 #define MP make_pair 29 30 #define REP(i,n) for(i=0;i<(n);++i) 31 #define FOR(i,l,h) for(i=(l);i<=(h);++i) 32 #define FORD(i,h,l) for(i=(h);i>=(l);--i) 33 34 typedef vector<int> VI; 35 typedef vector<string> VS; 36 typedef vector<double> VD; 37 typedef long long LL; 38 typedef pair<int,int> PII; 39 int fa[1200000]; 40 bool vis[1200000]; 41 42 class RightTriangle 43 { 44 public: 45 int find(int k){ 46 return !vis[k] ? k : fa[k] = find(fa[k]); 47 } 48 long long triangleCount(int n, int m, long long a, long long b, long long c) 49 { 50 if (n & 1) return 0; 51 memset(vis, 0, sizeof(vis)); 52 for (int i = 0; i < n; ++i) fa[i] = (i + 1) % n; 53 for (int i = 0; i < m; ++i) 54 vis[find((a * i * i + b*i +c) % n)] = true; 55 long long ans = 0; 56 for (int i = 0; i < n / 2; ++i) 57 if (vis[i] && vis[i + n / 2]) ++ ans; 58 return ans * (m - 2); 59 } 60 };
原文：http://www.cnblogs.com/yzcstc/p/3626927.html
