1、第一个，水题，直接求最大值即可，略。
2、根据题目，clipboard最大的时候是50，则可以得到，当粘贴超过50次以后，每次增长的个数是肯定是固定个数。
写的时候为了保险，把50换到了100多。
1 #include <iostream> 2 #include <string> 3 #include <vector> 4 #include <cstdlib> 5 #include <map> 6 #include <algorithm> 7 #include <stack> 8 #include <queue> 9 #include <cmath> 10 using namespace std; 11 typedef long long ll; 12 const int sz_pix = 160; 13 int num[sz_pix]; 14 bool judge[sz_pix][sz_pix]; 15 class PastingPaintingDivTwo { 16 public: 17 long long countColors(vector<string> clipboard, int T) { 18 int r = clipboard.size(); 19 int c = clipboard[0].size(); 20 for (int i = 0; i < sz_pix; i++) 21 for (int j = 0; j < sz_pix; j++) { 22 num[i] = 0; 23 judge[i][j] = false; 24 } 25 for (int i = 0; i < 105; i++) { 26 for (int a = 0; a < r; a++) { 27 for (int b = 0; b < c; b++) { 28 if ((!judge[i + a][i + b]) && (clipboard[a][b] == 'B')) { 29 judge[i + a][i + b] = true; 30 num[i]++; 31 } 32 } 33 } 34 } 35 if (T > 103) { 36 ll inc = num[102] ; 37 ll remain = T - 103; 38 ll tmp = remain * inc; 39 ll res = tmp; 40 for (int i = 0; i < 103; i++) { 41 res += num[i]; 42 } 43 return res; 44 } 45 ll res = 0; 46 for (int i = 0; i < T; i++) { 47 res += num[i]; 48 } 49 return res; 50 51 } 52 };
3、动态规划+状态压缩的问题。
1 #include <iostream> 2 #include <string> 3 #include <vector> 4 #include <cstdlib> 5 #include <map> 6 #include <algorithm> 7 #include <stack> 8 #include <queue> 9 #include <cmath> 10 using namespace std; 11 typedef long long ll; 12 ll num[15][15][1 << 14]; 13 14 bool path[15][15]; 15 class RandomOption { 16 public: 17 double getProbability(int keyCount, vector<int> badLane1, 18 vector<int> badLane2) { 19 20 for (int i = 0; i < keyCount; i++) 21 num[0][i][1 << i] = 1; 22 23 int sz = badLane1.size(); 24 for (int i = 0; i < sz; i++) { 25 int s = badLane1[i]; 26 int e = badLane2[i]; 27 path[s][e] = path[e][s] = true; 28 } 29 for (int i = 1; i < keyCount; i++) { 30 for (int j = 0; j < keyCount; j++) { 31 for (int pre = 0; pre < (1 << keyCount); pre++) { 32 for (int k = 0; k < keyCount; k++) { 33 if (path[j][k] || (pre & (1 << k))) 34 continue; 35 int now = pre | (1 << k); 36 num[i][k][now] += num[i - 1][j][pre]; 37 }; 38 } 39 } 40 41 } 42 double all = 1; 43 int keyCount1 = keyCount; 44 while (keyCount1) { 45 all *= keyCount1; 46 keyCount1--; 47 } 48 double pos = 0; 49 for (int i = 0; i < keyCount; i++) { 50 pos += num[keyCount - 1][i][(1 << keyCount) - 1]; 51 } 52 53 double res = pos / all; 54 return res; 55 } 56 };