感觉NOI题在向简单方向发展，或者说明年会难到暴呢？
直接模拟啊，枚举每个二进制数位，看经过变换之后是否为1及为1的条件即可。( O(nlogm))。
然后。。。跪了一个点，第五个死活比标准大一。。。
补码表示真dt，我会告诉你 1 >> 32 = 1吗（你肯定知道）？是我太傻逼了。
1 //{HEADS 2 #define FILE_IN_OUT 3 #include <cstdio> 4 #include <cstring> 5 #include <cstdlib> 6 #include <cmath> 7 #include <ctime> 8 #include <algorithm> 9 #include <iostream> 10 #include <fstream> 11 #include <vector> 12 #include <stack> 13 #include <queue> 14 #include <deque> 15 #include <map> 16 #include <set> 17 #include <bitset> 18 #include <complex> 19 #include <string> 20 #define REP(i, j) for (int i = 1; i <= j; ++i) 21 #define REPI(i, j, k) for (int i = j; i <= k; ++i) 22 #define REPD(i, j) for (int i = j; 0 < i; --i) 23 #define STLR(i, con) for (int i = 0, sz = con.size(); i < sz; ++i) 24 #define STLRD(i, con) for (int i = con.size() - 1; 0 <= i; --i) 25 #define CLR(s) memset(s, 0, sizeof s) 26 #define SET(s, v) memset(s, v, sizeof s) 27 #define mp make_pair 28 #define pb push_back 29 #define PL(k, n) for (int i = 1; i <= n; ++i) { cout << k[i] << ' '; } cout << endl 30 #define PS(k) STLR(i, k) { cout << k[i] << ' '; } cout << endl 31 using namespace std; 32 typedef long long LL; 33 typedef double DB; 34 typedef pair<int, int> i_pair; 35 const int INF = 0x3f3f3f3f; 36 //} 37 38 const int maxn = 1e5 + 100; 39 int ope[maxn], n, m; 40 bitset<33> t[maxn]; 41 char op[10]; 42 43 int me[40][2]; 44 45 int main() { 46 freopen("sleep.in", "r", stdin); 47 freopen("sleep.out", "w", stdout); 48 49 scanf("%d%d", &n, &m); 50 for(int i = 1; i <= n; ++i) { 51 int tmp; 52 scanf("%s %d", op, &tmp); 53 t[i] = tmp; 54 // printf("%s ", t[i].to_string().c_str()); 55 if(strcmp(op, "OR") == 0) { 56 ope[i] = 0; 57 } else if(strcmp(op, "AND") == 0) { 58 ope[i] = 1; 59 } else { 60 ope[i] = 2; 61 } 62 } 63 for(int i = 0; i <= 32; ++i) { 64 int init = 1; 65 for(int j = 1; j <= n; ++j) { 66 if(init == 1) { 67 if(ope[j] == 0) { 68 init = 1; 69 } else if(ope[j] == 1) { 70 if(t[j][i] == 0) { 71 init = 0; 72 } else if(t[j][i] == 1) { 73 init = 1; 74 } 75 } else { 76 if(t[j][i] == 0) { 77 init = 1; 78 } else if(t[j][i] == 1) { 79 init = 0; 80 } 81 } 82 } else if(init == 0) { 83 if(ope[j] == 0) { 84 if(t[j][i] == 1) { 85 init = 1; 86 } 87 } else if(ope[j] == 1) { 88 if(t[j][i] == 0) { 89 init = 0; 90 } else if(t[j][i] == 1) { 91 init = 0; 92 } 93 } else { 94 if(t[j][i] == 0) { 95 init = 0; 96 } else if(t[j][i] == 1) { 97 init = 1; 98 } 99 } 100 } 101 } 102 me[i][1] = init; 103 init = 0; 104 for(int j = 1; j <= n; ++j) { 105 if(init == 1) { 106 if(ope[j] == 0) { 107 init = 1; 108 } else if(ope[j] == 1) { 109 if(t[j][i] == 0) { 110 init = 0; 111 } else if(t[j][i] == 1) { 112 init = 1; 113 } 114 } else { 115 if(t[j][i] == 0) { 116 init = 1; 117 } else if(t[j][i] == 1) { 118 init = 0; 119 } 120 } 121 } else if(init == 0) { 122 if(ope[j] == 0) { 123 if(t[j][i] == 1) { 124 init = 1; 125 } 126 } else if(ope[j] == 1) { 127 if(t[j][i] == 0) { 128 init = 0; 129 } else if(t[j][i] == 1) { 130 init = 0; 131 } 132 } else { 133 if(t[j][i] == 0) { 134 init = 0; 135 } else if(t[j][i] == 1) { 136 init = 1; 137 } 138 } 139 } 140 } 141 me[i][0] = init; 142 } 143 int ans = 0; 144 int bit = 31; 145 /* 卧槽！！吧bit改成32就wa了一个点 146 * 出题人真良心没卡我 147 * 不然就10分了QAQ 148 * */ 149 bool free_flag = false; 150 for(int i = bit; 0 <= i; --i) { 151 if(free_flag) { 152 if(me[i][0] == 1 || me[i][1] == 1) { 153 ans += (1 << i); 154 } 155 continue; 156 } 157 if(me[i][0] == 1) { 158 ans += (1 << i); 159 if((m >> i) & 1) { 160 free_flag = true; 161 } 162 } else if(me[i][1] == 1) { 163 if((m >> i) & 1) { 164 ans += (1 << i); 165 } 166 } else if((m >> i) & 1) { 167 free_flag = true; 168 } 169 } 170 printf("%d ", ans); 171 fclose(stdin); 172 fclose(stdout); 173 return 0; 174 }