输入输出+暴力模拟入门：魔法之树、染色の树、矩阵、字母加密、玫瑰鸭

秋招实习刷题网站推荐：codefun2000.com，还有题解博客：blog.codefun2000.com/。以下内容都是来自塔子哥的~

输入输出

2023.04.15-春招-第三题-魔法之树

//#include
#include
#include

using namespace std;

typedef long long LL;
const int N = 1001;
LL n, l, r;
vector<int> weight(N);
vector<vector<int>> vec(N);//二维数组
// 图的存储:开一个全局的定长数组，其中每个元素都是一个不定长数组vector
// 开1001 是因为节点下标范围为[1,1000] , 所以需要多开一个
// 你所见到的开1005,1006也是这个原因。至少多开辟一位即可
int result = 0;

//传入当前节点 当前节点的父节点 累加的值 的值
void dfs(int u, int root, int pre)
{
    //获取遍历过路径 二进制 对应的十进制
    //获得当前权重 pre二进制进一位 相当于十进制*2
    //1--11  1--1*2+1

    int cur = pre * 2 + weight[u];
    cout << "pre = " << pre << "\tweight[u] = " << weight[u] << "\tcur = " << cur;

    //找到符合的路径
    if (cur > r) return;
    else if (cur >= l) result++;

    //遍历当前层
    for (auto& num : vec[u])
    {
        if (num == root) continue;//跳过父节点 一个父节点不能是权重
        //for(auto& vv : vec[u]) cout << "vv = " << vv << endl;
        cout << "\tnum = " << num << endl;
        dfs(num, u, cur);
    }
}

int main1()
{
    cin >> n >> l >> r;
    string str;
    cin >> str;
    for (int i = 1; i <= n; i++)
    {
        weight[i] = str[i-1] - '0';
    }

    // 由于是树,所以只需要读n - 1 条边
    // 由于你无法确认x,y之间谁是父亲节点，所以需要存双向边,
    // 在dfs的过程中防止返祖即可(返祖会引发死递归!)。
    // PS:在有一些题目里，他会明确规定谁是父亲,这种情况下就不用存双向边，
    // 且在dfs的过程中也不用担心返祖的问题.
    for (int i = 0; i < n - 1; i++)
    {
        int u, v;
        cin >> u >> v;
        vec[u].push_back(v);
        vec[v].push_back(u);
    }

    for (int i = 1; i <= n; i++)
    {
        //从父节点1开始出发 他没有父节点所以是-1 累加的十进制数是0
        cout << "i = " << i << endl;
        dfs(i, -1, 0);
    }
    cout << result << endl;

    //string str = "11010";
    //vector> nums = { {1,2,3}, {2,4,5 }, {1}, {2}, {2} };
    /*cout << "n = " << n << "\tl = " << l << "\tr = " << r << endl;
    cout << "str = " << str << endl;
    for (int i = 0; i < vec.size(); i++)
    {
        cout << "i = " << i << "\tvec[i] = " << vec[i][0] << endl;
        for (int j = i; j < vec[i].size(); j++)
        {
            cout << "\tvec[i][j] = " << vec[i][j] << endl;
        }
    }
    cout << "done" << endl;*/   
    system("pause");
    return 0;
}

2023.04.01-第五题-染色の树

//#include
#include 
#include 
using namespace std;
int n;
vector<vector<int>> edges;
vector<int> color;


int dfs(int root) {
    vector<int> tmp(2);//如果当前层有2个子节点 保存其两个子节点
    //如果    当前层有0个子节点    该节点的价值为    1
    //如果    当前层有1个子节点    该节点的价值为    其唯一子节点的价值
    //如果    当前层有2个子节点    该节点是红色，价值为两个子节点价值和；是绿色，价值为两个子节点价值的异或值
    
    if (edges[root].size() == 0) return 1;
    else if (edges[root].size() == 1) return dfs(edges[root][0]);
    else {
        for (int i = 0; i < 2; i++) 
            tmp[i] = dfs(edges[root][i]);//两个子节点的价值 为了计算父节点价值

        if (color[root - 1] == 1) {
            return tmp[0] + tmp[1];//父节点是红色
        }
        else return tmp[0] ^ tmp[1];//父节点是绿色
    }
}
int main() {
    cin >> n;
    if (n < 1) return 0;
    if (n == 1) return 1;
    color.resize(n + 1);
    edges.resize(n + 1);//4行
    //cout << edges.size() << endl;
    //保存     { {}, {2,3}, {}, {} }
    //过程是   p[i]=1 i从2开始 第2个父节点是1，i++；p[i]=1，i=3，第3个父节点是1
    int idx = 0;
    while (idx < n) {
        int tmp;
        cin >> tmp;//输入的是父节点
        edges[tmp].push_back(idx+2);//idx+2才是子节点 注意题目的i是从2开始
        idx++;
    }

    int idx2 = 0;
    while (idx2 < n) {
        int tmp;
        cin >> tmp;
        color[idx2++] = tmp;
    }
    /*for(int i = 0; i < edges[1].size(); i ++)
            cout << edges[1][i] << " ";*/
    //输入父节点值
    int res = dfs(1);
    //cout << res << endl;
    return 0;
}

取模模板

//#include
#include 
using namespace std;
const int mod = 1e9 + 7;

// -----取模操作模板----- 建议使用long long 实例化，最稳
template <typename T>
class Mod {
public:
	T add(T x, T y, T mod) {
		x %= mod;
		y %= mod;
		T res = (x + y) % mod;
		return res;
	}
	T sub(T x, T y, T mod) {
		x %= mod;
		y %= mod;
		T res = (x - y + mod) % mod;
		return res;
	}
	T mul(T x, T y, T mod) {
		x %= mod;
		y %= mod;
		T res = x * y % mod;
		return res;
	}
	T div(T x, T y, T mod) {
		x %= mod;
		y %= mod;
		T inv = fastPow(y, mod - 2, mod);
		T res = mul(x, inv, mod);
		return res;
	}
private:
	T fastPow(T a, T b, T mod) {
		T ans = 1, base = a;
		while (b) {
			if (b & 1) ans = mul(ans, base, mod);
			base = mul(base, base, mod);
			b >>= 1;
		}
		return ans;
	}
};
// -----取模操作模板 end-----
int main3() {
	int n;
	Mod<long long> t;
	cin >> n;
	for (int i = 1; i <= n; i++) {
		int op;
		long long x, y;
		cin >> op >> x >> y;
		if (op == 1) {
			cout << t.add(x, y, mod) << endl;
		}
		else if (op == 2) {
			cout << t.sub(x, y, mod) << endl;
		}
		else if (op == 3) {
			cout << t.mul(x, y, mod) << endl;
		}
		else {
			cout << t.div(x, y, mod) << endl;
		}
	}
}

暴力模拟入门

矩阵

#include
using namespace std;
int n;
const int mod = 1e9 + 7;
typedef long long LL;

template <typename T>
class MOD
{
public:
    T mul(T a, T b, T mod)
    {
        a %= mod;
        b %= mod;
        T res = (a * b) % mod;
        return res;
    }
    T fastPow(T a, T b, T mod)
    {
        T res = 1, base = a;
        while(b)
        {
            if((b & 1) == 1)
                mul(base, res, mod);//res *= a;
            mul(base, base, mod);//a *= a;
            b >>= 1;
        }
        return res;
    }
};

long long fast_pow(int a, int b)
{
    long long res = 1, temp = a;
    while(b)
    {
        if((b & 1) == 1)
            res *= temp;
        temp *= temp;
        b >>= 1;
    }
    return res;
}

int main()
{
    cin >> n;
    LL num = 0, result = 0;
    //num = n * (n-1) / 2;
    for(LL i=1; i<=n-1; i++)
        num += i;
    //cout << num << endl;

    //MOD mymod;
    //result = mymod.fastPow(2, num, mod);
    result = fast_pow(2, num);

    cout << result << endl;
    return 0;
}

字母加密

第一次写的时候没有考虑范围超出的时候

#include 
#include 
#include 
using namespace std;

string instr;
int main()
{
	cin >> instr;
	int len = instr.length();
	vector<int> a(len + 1);
	a[0] = 1, a[1] = 2, a[2] = 4;
	for (int i = 3; i < len + 1; i++)
	{
		a[i] = (a[i - 1] + a[i - 2] + a[i - 3]) % 26;
        //while (a[i] > 26) a[i] -= 26;
		//cout << a[i] << "  ";
	}
	string result;
	for (int i = 0; i < len; i++)
	{
		char temp = (instr[i] - 'a' + a[i]) % 26 + 'a';
		//cout << instr[i] - 'a' + a[i] << "  " << instr[i] + a[i] << "  ";
		//cout << (instr[i] - 'a' + a[i]) % 26<< "  " << temp << endl;
		result += temp;
	}
	cout << result << endl;
	return 0;
}

玫瑰鸭

#include
using namespace std;

typedef long long LL;
LL a, b, c;

int main()
{
	cin >> a >> b >> c;
	//LL a = 8;
	//LL b = 4;
	//LL c = 2;
	if (a > b) swap(a, b);//保证a是小的那个
	LL temp = b - a;//差值
	if (c - temp >= 0)//c比差值大 c补给a
	{
	//用c把差值补给a，让a = b
	a += temp;
	c -= temp;
	//如果c还有剩下 取c的一半，例如4 8 8
	a += c / 2;
	}
	else a += c;
	return a / 2;
}