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44 lines
1.1 KiB
44 lines
1.1 KiB
2 years ago
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## [$10117$. 「一本通 $4.1$ 练习 $2$」简单题](https://loj.ac/p/10117)
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#### 题目解析
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区间修改+单点查询,用树状数组维护差分数组，从而记录每个点反转的次数。最后单点查询点反转的次数%2即为应得值。
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### $Code$
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```cpp {.line-numbers}
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#include <bits/stdc++.h>
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using namespace std;
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const int N = 100005;
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int n, m;
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// 树状数组模板
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#define lowbit(x) (x & -x)
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typedef long long LL;
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int c[N];
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void add(int x, int v) {
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while (x < N) c[x] += v, x += lowbit(x);
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}
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LL sum(int x) {
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LL res = 0;
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while (x) res += c[x], x -= lowbit(x);
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return res;
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}
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int main() {
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scanf("%d%d", &n, &m);
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for (int i = 1; i <= m; i++) {
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int t, x, y;
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scanf("%d", &t);
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if (t == 1) {
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scanf("%d%d", &x, &y);
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add(x, 1), add(y + 1, -1); // 用差分数组+树状数组，模拟区间修改
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} else {
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scanf("%d", &x);
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printf("%d\n", sum(x) % 2); // 单点查询，是否变化
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}
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}
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return 0;
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}
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```
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