1、Calendar Game HDU - 1079
题意：两个游戏，Adam先手，每次将当前日期+1天或+1个月份（前提是加完后日期有效），谁先走到2001.11.4谁赢。现在给出一个起始日期（在1900.1.1~2001.11.4之间），问Adam能否赢。
思路：从最后一天2001.11.4向前枚举，确定必胜态和必败态。
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 using namespace std; 5 const int maxy = 2001; 6 const int maxm = 13; 7 const int maxd = 32; 8 int days[13] = { 0,31,28,31,30,31,30,31,31,30,31,30,31 }; 9 bool win[maxy][maxm][maxd]; 10 int cur_y = 2001, cur_m = 11, cur_d = 4; 11 bool isLeap(int year=cur_y) 12 { 13 if ((year % 100 != 0 && year % 4 == 0) || year % 400 == 0) return true; 14 else return false; 15 } 16 void sub() 17 { 18 cur_d--; 19 if (cur_d == 0) 20 { 21 cur_m--; 22 if (cur_m) cur_d = days[cur_m]; 23 else 24 { 25 cur_m = 12; 26 cur_y--; 27 if (isLeap()) days[2] = 29; 28 else days[2] = 28; 29 cur_d = days[cur_m]; 30 } 31 } 32 } 33 bool can_nextM() 34 { 35 int ty = cur_y, tm = cur_m; 36 tm++; 37 if (tm > 12) ty++, tm = 1; 38 if (ty > 2001) return false; 39 else if (ty == 2001) 40 { 41 if (tm > 11 || (tm == 11 && cur_d > 4)) return false; 42 else if (tm == 11 && cur_d <= 4) return true; 43 else 44 { 45 if (tm != 2) 46 { 47 if (cur_d > days[tm]) return false; 48 else return true; 49 } 50 else 51 { 52 int max_td = (isLeap(ty) ? 29 : 28); 53 if (cur_d > max_td) return false; 54 else return true; 55 } 56 } 57 } 58 else 59 { 60 if (tm != 2) 61 { 62 if (cur_d > days[tm]) return false; 63 else return true; 64 } 65 else 66 { 67 int max_td = (isLeap(ty) ? 29 : 28); 68 if (cur_d > max_td) return false; 69 else return true; 70 } 71 } 72 } 73 void solve() 74 { 75 bool pre=win[cur_y][cur_m][cur_d] = false; 76 while (cur_y != 1900 || cur_m != 1 || cur_d != 1) 77 { 78 sub(); 79 if (can_nextM()) 80 { 81 int ty = cur_y, tm = cur_m; 82 tm++; 83 if (tm > 12) ty++, tm = 1; 84 if (!pre || !win[ty][tm][cur_d]) pre = win[cur_y][cur_m][cur_d] = true; 85 else pre = win[cur_y][cur_m][cur_d] = false; 86 } 87 else 88 { 89 if (pre) pre = win[cur_y][cur_m][cur_d] = false; 90 else pre = win[cur_y][cur_m][cur_d] = true; 91 } 92 } 93 } 94 int main() 95 { 96 solve(); 97 int t; 98 scanf("%d", &t); 99 while (t--) 100 { 101 int y, m, d; 102 scanf("%d%d%d", &y, &m, &d); 103 if (win[y][m][d]) printf("YES "); 104 else printf("NO "); 105 } 106 return 0; 107 }