美团笔试题目

题目

给一个长度为n的数组，总权值和为val, 期望是否能找到一个区间[l,r](r - l + 1 < n)，使得区间[l, r]的权值和大于等于val，可能有某个元素值可以为负数。

为了省事，直接线段树暴力求某个区间的最大值。

#include<bits/stdc++.h>
using namespace std;
int N = 200005;
int tr[200015], f[50004], a[50004], add[200015];
void pushup(int root){
	tr[root] = max(tr[root << 1], tr[root << 1 | 1]);
}
void pushdown(int root){
	add[root << 1] += add[root];
	add[root << 1 | 1] += add[root];
	tr[root << 1] += add[root];
	tr[root << 1 | 1] += add[root];
	add[root] = 0;
}
void build(int l, int r, int root){
	if(l == r){
		tr[root] = f[l];
		return;
	}
	int mid = l + r >> 1;
	build(l, mid, root << 1);
	build(mid + 1, r , root << 1 | 1);
	pushup(root);
}
void undate(int l, int r, int root, int L, int R, int val){
	if(L <= l && R >= l){
		tr[root] += val;
		add[root] += val;
		return ;
	}
	pushdown(root);
	int mid = l + r >> 1;
	if(L <= mid) undate(l, mid, root << 1, L, R, val);
	if(R > mid) undate(mid + 1, r, root << 1 | 1, L, R, val);
	pushup(root);
}
int query(int l, int r, int root, int L, int R){
	if(L <= l && R >= r){
		return tr[root];
	}
	int ans = -1000000000, mid = l + r >> 1;
	pushdown(root);
	if(L <= mid) ans = max(ans, query(l, mid, root << 1, L, R));
	if(R > mid) ans = max(ans, query(mid + 1, r, root << 1 | 1, L, R));
	return ans;
}
int main(){
	int t, n;
	cin >> t;
	while(t--){
		cin >> n;
		f[0] = 0;
		int flag = 0;
		memset(tr, 0, sizeof(tr));
		memset(add, 0, sizeof(add));
		for(int i = 1; i <= n; i++){
			cin >> a[i];
			f[i] = a[i] + f[i - 1];
		}
		build(1, n - 1, 1);
		for(int i = 1; i <= n - 1; i++){
			int k = query(1, n - 1, 1, i, n - 1);
			if(k >= f[n]){
				flag = 1;
				break;
			}
			undate(1, n - 1, 1, i, n - 1, -a[i]);
		}
		int val = f[n];
		for(int i = 1; i <= n - 1; i++){
			f[n] -= a[i];
			if(f[n] >= val){
				flag = 1;
				break;
			}
		}
		flag == 1 ? cout << "Yes" << endl : cout << "No" << endl;
	}
    return 0;
}

求汉明距离

给两个01串\(s\)和\(t\),\(|t|\le |s|\), 想知道\(t\)和\(s\)中所有长度为\(t\)的子串的汉明距离之和
输入样例
01
00111
输出
3

#include<bits/stdc++.h>
using namespace std;
string t, s;
int main(){
	cin >> t >> s;
	int n = s.size();
	int m = t.size();
	bitset<50000> o1, o2, o3;
	int ans = 0, pos = 49999;
	for(int i = 0; i < m; i++){
		if(t[i] == '1') o2.set(pos - i);
	}
	for(int i = 0; i < m; i++){
		if(s[i] == '1') o1.set(pos - i);
	}
	o3 = o1 ^ o2;
	ans += o3.count();
	//cout << "o1" << " " << o1 << endl;
	//cout << "o2" << " " << o2 << endl;
	//cout << "o3" << " " << o3 << endl;
	//cout << "------------------------------" << endl;
	for(int i = m; i < n; i++){
		o1 <<= 1;
		if(s[i] == '1') o1.set(pos - m + 1);
		o3 = o1 ^ o2;
		ans += o3.count();
		//cout << "o1" << " " << o1 << endl;
		//cout << "o2" << " " << o2 << endl;
		//cout << "o3" << " " << o3 << endl;
		//cout << "------------------------------" << endl;
	}
	cout << ans << endl;
	return 0;
}