NOI2014 起床困难综合症 day1t1

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\n", 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\n", ans);
171     fclose(stdin);
172     fclose(stdout);
173     return 0;
174 }
巨丑无比的代码

 

posted @ 2014-07-29 18:15  sbit  阅读(354)  评论(0)  编辑  收藏  举报