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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 ith 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 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.
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 ith 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 |
Dataset is huge, use faster i/o methods.
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 }
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原文地址：http://www.cnblogs.com/zzuli2sjy/p/5890936.html
