SRM473

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:

 #line 7 "SequenceOfCommands.cpp"
 #include <cstdlib>
 #include <cctype>
 #include <cstring>
 #include <cstdio>
 #include <cmath>
 #include <algorithm>
 #include <vector>
 #include <string>
 #include <iostream>
 #include <sstream>
 #include <map>
 #include <set>
 #include <queue>
 #include <stack>
 #include <fstream>
 #include <numeric>
 #include <iomanip>
 #include <bitset>
 #include <list>
 #include <stdexcept>
 #include <functional>
 #include <utility>
 #include <ctime>
 using namespace std;

 #define PB push_back
 #define MP make_pair

 #define REP(i,n) for(i=0;i<(n);++i)
 #define FOR(i,l,h) for(i=(l);i<=(h);++i)
 #define FORD(i,h,l) for(i=(h);i>=(l);--i)

 typedef vector<int> VI;
 typedef vector<string> VS;
 typedef vector<double> VD;
 typedef long long LL;
 typedef pair<int,int> PII;
 const int g[][] = {{, },{,-},{-, },{, }};

 class SequenceOfCommands
 {
         public:
         string whatHappens(vector <string> commands)
         {
                 string S = accumulate(commands.begin(), commands.end(), string(""));
                 int x = , y = , dir = ;
                 int n = S.size();
                 for (int i = ; i < n; ++i){
                     if (S[i] == 'L') dir = (dir + ) % ;
                     if (S[i] == 'R') dir = (dir + ) % ;
                     if (S[i] == 'S'){
                          x += g[dir][];
                          y += g[dir][];
                     }
                 }
                 if (x ==  && y == ) return "bounded";
                 if (dir != ) return "bounded";
                 return "unbounded";
         }
 };

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:

 #line 7 "RightTriangle.cpp"
 #include <cstdlib>
 #include <cctype>
 #include <cstring>
 #include <cstdio>
 #include <cmath>
 #include <algorithm>
 #include <vector>
 #include <string>
 #include <iostream>
 #include <sstream>
 #include <map>
 #include <set>
 #include <queue>
 #include <stack>
 #include <fstream>
 #include <numeric>
 #include <iomanip>
 #include <bitset>
 #include <list>
 #include <stdexcept>
 #include <functional>
 #include <utility>
 #include <ctime>
 using namespace std;

 #define PB push_back
 #define MP make_pair

 #define REP(i,n) for(i=0;i<(n);++i)
 #define FOR(i,l,h) for(i=(l);i<=(h);++i)
 #define FORD(i,h,l) for(i=(h);i>=(l);--i)

 typedef vector<int> VI;
 typedef vector<string> VS;
 typedef vector<double> VD;
 typedef long long LL;
 typedef pair<int,int> PII;
 int fa[];
 bool vis[];

 class RightTriangle
 {
         public:
         int find(int k){
             return !vis[k] ? k : fa[k] = find(fa[k]);
         }
         long long triangleCount(int n, int m, long long a, long long b, long long c)
         {
                if (n & ) return ;
                memset(vis, , sizeof(vis));
                for (int i = ; i < n; ++i) fa[i] = (i + ) % n;
                for (int i = ; i < m; ++i)
                    vis[find((a * i * i + b*i +c) % n)] = true;
                long long ans = ;
                for (int i = ; i < n / ; ++i)
                    if (vis[i] && vis[i + n / ]) ++ ans;
                return ans * (m - );
         }
 };