洛谷P5057简单题

P5057 [CQOI2006] 简单题

这是题面

思路

每次操作，直接区间加\(1\)，最后求结果的时候对\(2\)取余就好了

这个题就是区间修改 + 单点查询

可以用树状数组或者线段数维护

代码

#include <bits/stdc++.h>
#define endl '\n'
#define int long long
#define lowbit(x) x & -x
const int maxn = 5e5 + 5;
const int inf = 0x7f7f7f7f;struct custom_hash 
{static uint64_t splitmix64(uint64_t x) {x ^= x << 13;x ^= x >> 7;x ^= x << 17;return x; }size_t operator () (uint64_t x) const {static const uint64_t FIXED_RANDOM = std::chrono::steady_clock::now().time_since_epoch().count(); // 时间戳return splitmix64(x + FIXED_RANDOM);}
};
int n = 0, m = 0;
int fenwick1[maxn], fenwick2[maxn];void modify(int pos, int x)
{int v = pos * x;while(pos <= n){fenwick1[pos] += x;fenwick2[pos] += v;pos += lowbit(pos);}
}   int query(int pos, int t[])
{int res = 0;while(pos){res += t[pos];pos -= lowbit(pos);}return res;
}void range_modify(int l, int r, int v)
{modify(l, v);modify(r + 1, -v);
}int range_query(int l, int r)
{int pre = (r + 1) * query(r, fenwick1) - l * query(l - 1, fenwick1);int pos = query(r, fenwick2) - query(l - 1, fenwick2);return pre - pos;
}int single_query(int pos)
{return query(pos, fenwick1);
}void solve()
{std::cin >> n >> m;int op = 0, L = 0, R = 0;for (int i = 1; i <= m; i++){std::cin >> op;if (op == 1){std::cin >> L >> R;range_modify(L, R, 1);}else{std::cin >> L;int ans = single_query(L);std::cout << ans % 2 << endl;}}
}signed main()
{std::ios::sync_with_stdio(false);std::cin.tie(nullptr); std::cout.tie(nullptr);//freopen("out.txt", "w", stdout);int t = 1;//std::cin >> t;while(t--){solve();}return 0;
}