lightoj 1080 Binary Simulation (线段树区间更新)

1080 - Binary Simulation

PDF (English)StatisticsForum

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 and1 ≤ 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

题目链接：http://lightoj.com/volume_showproblem.php?problem=1080

题目大意：两个操作I x y把[x,y]内的数字取反，Q x查询第x位的数字是0还是1

题目分析：因为同一个区间取反偶数次和原数相同，因此只需要累计区间取反次数即可，用线段树维护，区间更新

#include <cstdio>#include <cstring>#include <algorithm>#define lson l, mid, rt << 1#define rson mid + 1, r, rt << 1 | 1using namespace std;int const MAX = 1e5 + 5;int chg[MAX << 2], lazy[MAX << 2];char s[MAX];int n, len;void PushDown(int rt){if(lazy[rt]){chg[rt << 1] += lazy[rt];chg[rt << 1 | 1] += lazy[rt];lazy[rt << 1] += lazy[rt];lazy[rt << 1 | 1] += lazy[rt];lazy[rt] = 0;}return;}void Update(int L, int R, int l, int r, int rt){if(L <= l && r <= R){chg[rt] ++;lazy[rt] ++;return;}int mid = (l + r) >> 1;PushDown(rt);if(L <= mid)Update(L, R, lson);if(mid < R)Update(L, R, rson);}int Query(int L, int R, int l, int r, int rt){if(L == l && r == R)return chg[rt];int mid = (l + r) >> 1, ans = 0;;PushDown(rt);if(L <= mid)return Query(L, R, lson);if(mid < R)return Query(L, R, rson);return ans;}int main(){int T;scanf("%d", &T);for(int ca = 1; ca <= T; ca++){scanf("%s %d", s + 1, &n);len = strlen(s + 1);memset(chg, 0, sizeof(chg));memset(lazy, 0, sizeof(lazy));printf("Case %d:\n", ca);while(n --){char op[2];scanf("%s", op);if(op[0] == 'I'){int l, r;scanf("%d %d", &l, &r);Update(l, r, 1, len, 1);}else{int x, num = 0;scanf("%d", &x);num = Query(x, x, 1, len, 1);if(num & 1)printf("%d\n", (s[x] == '0') ? 1 : 0);elseprintf("%d\n", (s[x] == '0') ? 0 : 1);}}}}