线段树 Codeforces Round #197 (Div. 2) D. Xenia and Bit Operations

 

题目传送门

 1 /*
 2     线段树的单点更新：有一个交叉更新，若rank=1，or；rank=0，xor
 3     详细解释：http://www.xuebuyuan.com/1154895.html
 4 */
 5 #include 
 6 #include 
 7 #include 
 8 #include 
 9 #include <string>
10 #include 
11 #include <set>
12 #include );
77 
78     int p, b;
79     for (int i=1; i<=m; ++i)
80     {
81         scanf ("%d%d", &p, &b);
82         updata (p, b, 1, 1<1);
83         printf ("%d\n", node[1].v);
84     }
85 
86     return 0;
87 }

 

转载于:https://www.cnblogs.com/Running-Time/p/4402189.html