1080 - Binary Simulation[树状数组]

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.

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PROBLEM SETTER: JANE ALAM JAN

题目大意是这样的，给你一个长度为n的01字串，其实下标为1，有两种操作：

I a b 代表将区间为[a,b]的串的0变成1,1变成0，也就是说反转

Q i，代表查询下标为i的数字

总结下来就是 区间更新，单点查询问题，可以用线段树做，也可以用树状数组做。

那么我们记录下来这一个点被反转了多少次，然后用当前的位置的数字加上反转次数模2就是查询的答案。

我们这样，开一个长度为n的整形数组s，在执行I操作时，把s[a]加1，把s[b+1]减1

那么我们如果想知道i点被反转了多少次，就计算s[1]+s[2]+s[3]+...+s[i]，这就是反转了多少次。

由于是前缀和，我们可以用树状数组维护

代码如下：

#include <bits/stdc++.h>
using namespace std;
 
const int MAX_N = 100100;
int bit[MAX_N],n,T,q;
char s[MAX_N];
 
void add(int i,int x){
    while(i<=n){
        bit[i] += x;
        i += i&-i;
    }
}
int sum(int i){
    int s = 0;
    while(i>0){
        s += bit[i];
        i -= i&-i;
    }
    return s;
}
 
int main(){
    scanf("%d",&T);
    for(int t=1;t<=T;t++){
        printf("Case %d:\n",t);
        scanf("%s",s);
        memset(bit,0,sizeof(bit));
        n = strlen(s);
        scanf("%d",&q);
        while(q--){
            char cmd[2];
            scanf("%s",cmd);
            if(cmd[0]=='I'){
                int a,b;
                scanf("%d%d",&a,&b);
                add(a,1);
                add(b+1,-1);
            } else {
                int pos;
                scanf("%d",&pos);
                int res = sum(pos);
                int ans = s[pos-1]-'0'+res;
                printf("%d\n",ans%2);
            }
        }
    }
    return 0;
}