模拟赛#2 | 牛客普及周赛

题目链接:https://ac.nowcoder.com/acm/contest/59457#rank

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AB 纯暴力
C
思维题
从任意情况入手, 设\(highbit(x)\)为数x的二进制表示最高位, 则容易发现当\(highbit(r)\)大于\(highbit(l)\)时, 总能异或出从\(highbit(r)\)到第一位每一位都为1的最优状况, 继续考虑当\(highbit(r) = highbit(l)\)时, 最后一位无论如何都不能有变化, 考虑下一位, 假如相同时情况类似, 继续考虑下一位... 容易猜想到需要找值不同的最高位来计算答案

(这题70分特别好拿, 两个特判加暴力就70了, 比赛的时候5分钟就拿了70分, 然后剩下30分想了一个小时)

代码

/*
Author: SJ
*/
#include<bits/stdc++.h>
const int N = 1e5 + 10;
using ll = long long;
using ull = unsigned long long;
 
ll t, l, r, ans;
int lowbit(ll x) {
    return x & -x;
}
int find(ll x) {
    int pos = 0;
    while (x) {
        pos++;
        x >>= 1;
    }
    return pos;
}
int find2(ll x, ll y) {
    int pos = 0, cnt = 0;
    while (x && y) {
        cnt++;
        int tmp1 = (x & 1);
        int tmp2 = (y & 1);
        if (tmp1 != tmp2) {
            pos = cnt;
        }
        x >>= 1;
        y >>= 1;
    }
    return pos;
}
void solve() {
    std::cin >> l >> r;
    ans = 0;
    if (l == r) {
        std::cout << 0 << "\n";;
        return;
    } 
    int pos1 = find(l), pos2 = find(r);
    if (pos2 > pos1) {
        std::cout << (ll)std::pow(2, pos2) - 1 << "\n";
        return;
    }
    int pos3 = find2(l, r);
    std::cout << (ll)std::pow(2, pos3) - 1 << "\n";
}
int main() {
    std::ios::sync_with_stdio(false);
    std::cin.tie(nullptr);
 
    std::cin >> t;
    while (t--) solve();
    return 0;
}

D
归并排序 + 数学
先用归并排序\(O(nlogn)\)求出初始串逆序对的奇偶性, 然后根据"排列中任意两个数对换奇偶性改变"这一结论来\(O(1)\)回答即可

代码

/*
Author: SJ
*/
#include<bits/stdc++.h>
const int N = 1e5 + 10;
using ll = long long;
using ull = unsigned long long;

ll n, m, t[N], a[N], cnt, mark1;
void merge_sort(ll a[], int l, int r) {
	if (l == r) return;
	int mid = (l + r) / 2;
	merge_sort(a, l, mid), merge_sort(a, mid + 1, r);
	for (int i = l, j = l, k = mid + 1; i <= r; i++) {
		if (j == mid + 1) {
			t[i] = a[k++];
		} else if (k == r + 1) {
			cnt += k - mid - 1;
			t[i] = a[j++];
		} else if (a[j] <= a[k]) {
			cnt += k - mid - 1;
			t[i] = a[j++];
		} else {
			t[i] = a[k++];
		}
	}
	for (int i = l; i <= r; i++) {
		a[i] = t[i];
	}
}
int main() {
	std::ios::sync_with_stdio(false);
	std::cin.tie(nullptr);

	std::cin >> n;
	for (int i = 1; i <= n; i++) {
		std::cin >> a[i];
	}
	merge_sort(a, 1, n);
	if (cnt % 2) mark1 = 1;
	else mark1 = 0;


	std::cin >> m;
	for (int i = 1; i <= m; i++) {
		int l, r;
		std::cin >> l >> r;
		if ((r - l + 1) % 2 == 0) {
			int check1 = (r - l + 1) / 2;
			mark1 = (check1 % 2) ? ((mark1 + 1) % 2) : mark1;
		} else {
			int check1 = (r - l) / 2;
			mark1 = (check1 % 2) ? ((mark1 + 1) % 2) : mark1;
		}
		if (mark1 == 0) std::cout << "like" << "\n";
		else std::cout << "dislike" << "\n";
	}
	return 0;
}

期望得分: 400/400

C这种题知道结论就一眼, 真是虚头巴脑