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TangDou/AcWing/BIT/LOJ10117.md

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+## [$10117$. 「一本通 $4.1$ 练习 $2$」简单题](https://loj.ac/p/10117)
+
+#### 题目解析
+
+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。
+
+### $Code$
+```cpp {.line-numbers}
+#include <bits/stdc++.h>
+using namespace std;
+const int N = 100005;
+int n, m;
+
+// 树状数组模板
+#define lowbit(x) (x & -x)
+typedef long long LL;
+int c[N];
+void add(int x, int v) {
+    while (x < N) c[x] += v, x += lowbit(x);
+}
+LL sum(int x) {
+    LL res = 0;
+    while (x) res += c[x], x -= lowbit(x);
+    return res;
+}
+
+int main() {
+    scanf("%d%d", &n, &m);
+    for (int i = 1; i <= m; i++) {
+        int t, x, y;
+        scanf("%d", &t);
+        if (t == 1) {
+            scanf("%d%d", &x, &y);
+            add(x, 1), add(y + 1, -1); // 用差分数组+树状数组，模拟区间修改
+        } else {
+            scanf("%d", &x);
+            printf("%d\n", sum(x) % 2); // 单点查询，是否变化
+        }
+    }
+    return 0;
+}
+
+```