洛谷P5057簡單題

P5057 [CQOI2006] 簡單題

這是題面

思路

每次操作，直接區間加\(1\)，最後求結果的時候對\(2\)取餘就好了

這個題就是區間修改 + 單點查詢

可以用樹狀陣列或者線段數維護

程式碼

#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(); // 時間戳
		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;
}