SRM482

250pt

题意：给定n把锁，第i轮每间隔i个打开一个木有打开的。问最后打开的事几

思路：直接vector模拟

code：

 #line 7 "LockersDivOne.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;

 class LockersDivOne
 {
         public:
         int get(int n){
               vector<int> a, b;
               for (int i = ; i < n; ++i)
                   a.push_back(i);
               int res = ;
               for (int i = ; ; ++i){
                     if (a.size() == ) break;
                     b.clear();
                     for (int j = ; j < a.size(); j++)
                        if (j % i == ) res = a[j] + ;
                        else b.push_back(a[j]);
                     a = b;
               }
               return res;

         }
         int lastOpened(int N)
         {
                 return get(N);
         }
 };

500pt

题意：A和B两个人玩汉诺塔，其中A移动汉诺塔用最快的方法，移动的步数是2^n-1步，而B用的方法可以保证每种状态恰好被访问到一次，移动的步数是3^n-1。两个人移动的伪代码都给定，问第一个人移动K步后的configuration，按照第二个人的方法需要移动多少步。

思路:先算出最终的状态。然后根据最终的状态用递归算出答案。

code:

 #line 7 "HanoiGoodAndBad.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 thr[];

 class HanoiGoodAndBad
 {
         public:
         int dd, ans;
         int H[][];
         int top[];
         void solve_Dave(int a, int c, int b, int n){
             if(n > ){
                 if (dd == ) return;
                 solve_Dave(a, b, c, n-);
                 if (dd == ) return;
                 H[c][++top[c]] = n;
                 H[a][top[a]--] = ;
                 --dd;
                 solve_Dave(b, c, a, n-);
             }
         }
         void solve_Earl(int a, int c, int b, int n){
               if (n == ) return;
               int P = ;
               if (H[][top[]] == n) P = ;
               if (H[][top[]] == n) P = ;
               top[P]--;
               if (P == c){
                       ans +=  * thr[n-];
                       solve_Earl(a, c, b, n-);
               }
               if (P == b){
                       ans += thr[n-];
                       solve_Earl(c, a, b, n-);
               }
               if (P == a) solve_Earl(a, c, b, n-);
         }
         int moves(int N, int Dave)
         {
              dd = Dave;
              memset(top, , sizeof(top));
              memset(H, , sizeof(H));
              for (int i = N; i >= ; --i)  H[][++top[]] = i;
              solve_Dave(, , , N);
              ans = ;
              for (int i = ; i < ; ++i)
                  for (int j = ; j <= top[i]/ ; ++j)
                     swap(H[i][j], H[i][top[i]-j+]);
              thr[] = ;
              for (int i = ; i <= ; ++i) thr[i] = thr[i-] * ;
              solve_Earl(, , , N);
              return ans;
         }
 };