1080 - Binary Simulation

Posted 2020-08-09 SJY

1080 - Binary Simulation

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Time Limit: 2 second(s)

Memory Limit: 64 MB

Given a binary number, we are about to do some operations on the number. Two types of operations can be here.

‘I i j‘    which means invert the bit from i to j (inclusive)

‘Q i‘    answer whether the i^th bit is 0 or 1

The MSB (most significant bit) is the first bit (i.e. i=1). The binary number can contain leading zeroes.

Input

Input starts with an integer T (≤ 10), denoting the number of test cases.

Each case starts with a line containing a binary integer having length n (1 ≤ n ≤ 10⁵). The next line will contain an integer q (1 ≤ q ≤ 50000) denoting the number of queries. Each query will be either in the form ‘I i j‘ where i, j are integers and 1 ≤ i ≤ j ≤ n. Or the query will be in the form ‘Q i‘ where i is an integer and 1 ≤ i ≤ n.

Output

For each case, print the case number in a single line. Then for each query ‘Q i‘ you have to print 1 or 0 depending on the i^th bit.

Sample Input	Output for Sample Input
2 0011001100 6 I 1 10 I 2 7 Q 2 Q 1 Q 7 Q 5 1011110111 6 I 1 10 I 2 7 Q 2 Q 1 Q 7 Q 5	Case 1: 0 1 1 0 Case 2: 0 0 0 1

Note

Dataset is huge, use faster i/o methods.

---

PROBLEM SETTER: JANE ALAM JAN

思路：线段树；

线段树维护加反了多少次。

 1 #include<stdio.h>
 2 #include<iostream>
 3 #include<algorithm>
 4 #include<string.h>
 5 #include<queue>
 6 #include<math.h>
 7 using namespace std;
 8 char str[100005];
 9 int tree[4*100005];
10 void in(int l,int r,int k,int nn,int mm)
11 {
12     if(l>mm||r<nn)
13     {
14         return ;
15     }
16     else if(l<=nn&&r>=mm)
17     {
18         tree[k]++;
19         tree[k]%=2;
20         return ;
21     }
22     else
23     {
24         tree[2*k+1]+=tree[k];
25         tree[2*k+1]%=2;
26         tree[2*k+2]+=tree[k];
27         tree[2*k+2]%=2;
28         tree[k] = 0;
29         in(l,r,2*k+1,nn,(nn+mm)/2);
30         in(l,r,2*k+2,(nn+mm)/2+1,mm);
31     }
32 }
33 int ask(int l,int r,int k,int nn,int mm)
34 {
35     if(l>mm||r<nn)
36     {
37         return 0;
38     }
39     else if(l<=nn&&r>=mm)
40     {
41         return tree[k];
42     }
43     else
44     {
45         tree[2*k+1]+=tree[k];
46         tree[2*k+1]%=2;
47         tree[2*k+2]+=tree[k];
48         tree[2*k+2]%=2;
49         tree[k] = 0;
50         int nx = ask(l,r,2*k+1,nn,(nn+mm)/2);
51         int ny = ask(l,r,2*k+2,(nn+mm)/2+1,mm);
52         return (nx + ny)%2;
53     }
54 }
55 int main(void)
56 {
57     int T;
58     scanf("%d",&T);
59     int __ca = 0;
60     while(T--)
61     {
62         __ca++;
63         printf("Case %d:\n",__ca);
64         memset(tree,0,sizeof(tree));
65         scanf("%s",str);
66         int n;int l = strlen(str);
67         scanf("%d ",&n);
68         while(n--)
69         {
70             char a[10];
71             int x,y;
72             scanf("%s",a);
73             if(a[0] == ‘I‘)
74             {scanf("%d %d",&x,&y);
75              in(x-1,y-1,0,0,l-1);
76             }
77             else
78             {
79                 int x;
80                 scanf("%d",&x);
81                 int ny = ask(x-1,x-1,0,0,l-1);
82                 printf("%d\n",(str[x-1]-‘0‘+ny)%2);
83             }
84         }
85     }return 0;
86 }