华为笔试（9月20日）

T1 100% 一个循环数组 如果直接就是递增的就不用处理 否则需要找个递减点先把他按顺序弄回去 然后计算起始结束下标要把移动的下标计算进去

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

int main() {
    int n, target;
    cin >> n;
    vector<int> a(n);
    for (int i = 0; i < n; i++) {
        cin >> a[i];
    }
    cin >> target;

    int change = 0;
    for (int i = 0; i < n - 1; i++) {
        if (a[i] > a[i + 1]) {
            change = i + 1;
            vector<int> tmp(n);
            for (int j = 0; j < n; j++) {
                tmp[j] = a[(j + change) % n];
            }
            a = tmp;
            break;
        }
    }

    int start = INT_MAX;
    int end = INT_MIN;
    for (int i = 0; i < n; i++) {
        if (a[i] == target) {
            start = min(start, i);
            end = max(end, i);
        }
    }
    cout << (start + change) % n << " " << (end + change) % n << endl;
    return 0;
}

T2 100% 将相邻的1连起来，然后利用多源bfs分层，首先将最左侧是1压入队列中，跑到最右侧，层数最小的就是最短的路径

#include<iostream>
#include<vector>
#include<queue>
using namespace std;


bool ifon(int x, int y, int n, int m)
{
	return x >= 0 && x < n&& y >= 0 && y < m;
}

int main()
{
	int n, m;
	cin >> n >> m;
	vector<vector<int>> grid(n, vector<int>(m, 0));
	auto vis = grid;
	for (int i = 0; i < n; i++)
		for (int j = 0; j < m; j++)
			cin >> grid[i][j];
	queue<pair<int,int>> q;
	queue<int> sq;
	for (int i = 0; i < n; i++) {
		if (grid[i][0] == 1) {
			q.push({ i,0 });
			vis[i][0] = 1;
			sq.push(0);
		}
	}
	int ans = 1e9;
	int dx[4] = {-1,1,0,0};
	int dy[4] = { 0,0,-1,1 };
	while (q.size())
	{
		int x = q.front().first, y = q.front().second;
		int stp = sq.front();
		q.pop();
		sq.pop();
		if (y == m - 1)
		{
			ans = min(ans, stp);
			continue;
		}
		for (int i = 0; i < 4; i++)
		{
			int tox = x + dx[i], toy = y + dy[i];
			if (ifon(tox, toy, n, m) == 0)
				continue;
			if (vis[tox][toy]||grid[tox][toy]==0)continue;
			vis[tox][toy] = 1;
			q.push({ tox,toy });
			sq.push(stp + 1);
		}
	}
	if (ans == 1e9)
		cout << -1;
	else
		cout << ans;
}

T3 65% 大模拟，用map<string,int> mp存储每个变量的值，然后利用字符串切割处理字符串，模拟计算，注意溢出用long long

#include<iostream>
#include<vector>
#include<string>
#include<map>
using namespace std;

long long cau(long long a, char c, long long b)
{
	if (c == '+') return a + b;
	else if (c == '-') return a - b;
	else if (c == '*') return a * b;
	else return a / b;
}

int main()
{
	map<string, long long> mp;
	string  op;
	cin >> op;
	while (1)
	{
		if (op == "let")
		{
			string name;
			cin >> name;
			int right = 1;
			if (name[0] >= '0' && name[0] <= '9')
			{
				right = 0;
			}
			for (char c : name)
			{
				if (!(c == '_' || (c >= '0' && c <= '9') ||
					(c >= 'a' && c <= 'z') || (c >= 'A' && c <= 'Z')))
					right = 0;
			}
			string next;
			cin >> next;
			if (next != "=")
				right = 0;
			if (right == 0)
			{
				cout << "<syntax-error>" << endl;
				return 0;
			}
			string pre;
			cin >> pre;
			if (pre[0] == '-' || (pre[0] >= '0' && pre[0] <= '9'))
				mp[name] = stoll(pre.c_str());
			else
				mp[name] = mp[pre];

			string pa;

			while (cin >> pa)
			{
				if (pa == "let" || (pa.size() >= 3 && pa.substr(0, 3) == "out"))
					break;
				if (pa[0] == '+' || pa[0] == '-' || pa[0] == '*' || pa[0] == '/') {
					pre = pa;
					continue;
				}

				if (pa[0] == '-' || (pa[0] >= '0' && pa[0] <= '9'))
					mp[name] = cau(mp[name], pre[0], stoll(pa.c_str()));
				else
					mp[name] = cau(mp[name], pre[0], mp[pa]);
			}
			op = pa;
		}
		else if (op.size() >= 3 && op.substr(0, 3) == "out")
		{
			int n = op.size();
			string pa = op.substr(4, n - 5);
			long long pt = 0;
			if (pa[0] == '-' || (pa[0] >= '0' && pa[0] <= '9'))
				pt = stoll(pa.c_str());
			else if (mp.find(pa) != mp.end())
				pt = mp[pa];
			if (pt < -2147483648)
				cout << "<underflow>" << endl;
			else if (pt > 2147483647)
				cout << "<overflow>" << endl;
			else if (mp.find(pa) == mp.end())
				cout << "<undefined>" << endl;
			else
				cout << pt << endl;
			op = "over";
			cin >> op;
		}
		else
			break;
	}
}