import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt(), m = sc.nextInt();
        String s = sc.next();
        while (m-- > 0) {
            int l = sc.nextInt(), r = sc.nextInt();
            String[] c = sc.nextLine().trim().split("\\s");
            String tmp = s.substring(l - 1, r).replaceAll(c[0], c[1]);
            s = s.substring(0, l - 1) + tmp + s.substring(r);
        }
        System.out.println(s);
    }
}

水一个吧

class Solution {
public:
    vector<bool> st;
    vector<int> path;
    vector<vector<int>> ans;

    vector<vector<int>> permutation(vector<int>& nums) {
        sort(nums.begin(), nums.end());
        st = vector<bool>(nums.size(), false);
        path = vector<int>(nums.size());
        dfs(nums, 0, 0);
        return ans;
    }

    void dfs(vector<int>& nums, int u, int start)
    {
        if (u == nums.size())
        {
            ans.push_back(path);
            return;
        }

        for (int i = start; i < nums.size(); i ++ )
            if (!st[i])
            {
                st[i] = true;
                path[i] = nums[u];
                if (u + 1 < nums.size() && nums[u + 1] != nums[u])
                    dfs(nums, u + 1, 0);
                else
                    dfs(nums, u + 1, i + 1);
                st[i] = false;
            }
    }

};

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<vector<int>> printFromTopToBottom(TreeNode* root) {
        vector<vector<int>>res;
        if(!root) return res;
        queue<TreeNode*>q;
        q.push(root);

        int cnt=1;
        while(q.size())
        {
            int len=q.size();   
            vector<int>level;
            for(int i=0;i<len;i++)
            {
                auto t=q.front();
                q.pop();

                level.push_back(t->val);
                if(t->left)q.push(t->left);
                if(t->right) q.push(t->right);
            }
            if(cnt%2==0) reverse(level.begin(),level.end());
            res.push_back(level);
            cnt++;
        }
        return res;
    }
};

偶数层的时候翻转一下就可以了

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<vector<int>> printFromTopToBottom(TreeNode* root) {
        vector<vector<int>>res;
        if(!root) return res;
        queue<TreeNode*>q;
        q.push(root);

        int cnt=1;
        while(q.size())
        {
            int len=q.size();   
            vector<int>level;
            for(int i=0;i<len;i++)
            {
                auto t=q.front();
                q.pop();

                level.push_back(t->val);
                if(t->left)q.push(t->left);
                if(t->right) q.push(t->right);
            }
            if(cnt%2==0) reverse(level.begin(),level.end());
            res.push_back(level);
            cnt++;
        }
        return res;
    }
};

我们再每一层结束的时候，往queue中加入一个null标记
在queue中读取出一个数，如果是标记，看看整个树是否遍历完了(如果完了直接退出)，如果没有就将其放入level放入res中并清空level的状态在queue中放入nullptr

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<vector<int>> printFromTopToBottom(TreeNode* root) {
        vector<vector<int>>res;
        if(!root) return res;

        queue<TreeNode*>q;
        q.push(root);
        q.push(nullptr);

        vector<int>level;
        while(q.size())
        {
            auto t=q.front();
            q.pop();
            if(!t)
            {
                if(level.empty()) break;
                res.push_back(level);
                level.clear();
                q.push(nullptr);
                continue;
            }
            level.push_back(t->val);
            if(t->left) q.push(t->left);
            if(t->right) q.push(t->right);
        }
        return res;
    }
};

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<vector<int>> printFromTopToBottom(TreeNode* root) {
        vector<vector<int>>res;
        if(!root) return res;

        queue<TreeNode*>q;
        q.push(root);
        q.push(nullptr);

        vector<int>level;
        while(q.size())
        {
            auto t=q.front();
            q.pop();
            if(!t)
            {
                if(level.empty()) break;
                res.push_back(level);
                level.clear();
                q.push(nullptr);
                continue;
            }
            level.push_back(t->val);
            if(t->left) q.push(t->left);
            if(t->right) q.push(t->right);
        }
        return res;
    }
};

#include <iostream>
#include <algorithm>
#include <vector>

using namespace std;


const int N = 5010;

int primes[N], cnt;
int sum[N];
bool st[N];


void get_primes(int n)
{
    for (int i = 2; i <= n; i ++ )
    {
        if (!st[i]) primes[cnt ++ ] = i;
        for (int j = 0; primes[j] <= n / i; j ++ )
        {
            st[primes[j] * i] = true;
            if (i % primes[j] == 0) break;
        }
    }
}


int get(int n, int p)
{
    int res = 0;
    while (n)
    {
        res += n / p;
        n /= p;
    }
    return res;
}


vector<int> mul(vector<int> a, int b)
{
    vector<int> c;
    int t = 0;
    for (int i = 0; i < a.size(); i ++ )
    {
        t += a[i] * b;
        c.push_back(t % 10);
        t /= 10;
    }
    while (t)
    {
        c.push_back(t % 10);
        t /= 10;
    }
    return c;
}


int main()
{
    int a, b;
    cin >> a >> b;

    get_primes(a);

    for (int i = 0; i < cnt; i ++ )
    {
        int p = primes[i];
        sum[i] = get(a, p) - get(a - b, p) - get(b, p);
    }

    vector<int> res;
    res.push_back(1);

    for (int i = 0; i < cnt; i ++ )
        for (int j = 0; j < sum[i]; j ++ )
            res = mul(res, primes[i]);

    for (int i = res.size() - 1; i >= 0; i -- ) printf("%d", res[i]);
    puts("");

    return 0;
}

#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>

using namespace std;

const int N = 110;
const double eps = 1e-8;

int n;
double a[N][N];

int gauss()  // 高斯消元，答案存于a[i][n]中，0 <= i < n
{
    int c, r;
    for (c = 0, r = 0; c < n; c ++ )
    {
        int t = r;
        for (int i = r; i < n; i ++ )  // 找绝对值最大的行
            if (fabs(a[i][c]) > fabs(a[t][c]))
                t = i;

        if (fabs(a[t][c]) < eps) continue;

        for (int i = c; i <= n; i ++ ) swap(a[t][i], a[r][i]);  // 将绝对值最大的行换到最顶端
        for (int i = n; i >= c; i -- ) a[r][i] /= a[r][c];  // 将当前行的首位变成1
        for (int i = r + 1; i < n; i ++ )  // 用当前行将下面所有的列消成0
            if (fabs(a[i][c]) > eps)
                for (int j = n; j >= c; j -- )
                    a[i][j] -= a[r][j] * a[i][c];

        r ++ ;
    }

    if (r < n)
    {
        for (int i = r; i < n; i ++ )
            if (fabs(a[i][n]) > eps)
                return 2; // 无解
        return 1; // 有无穷多组解
    }

    for (int i = n - 1; i >= 0; i -- )
        for (int j = i + 1; j < n; j ++ )
            a[i][n] -= a[i][j] * a[j][n];

    return 0; // 有唯一解
}


int main()
{
    scanf("%d", &n);
    for (int i = 0; i < n; i ++ )
        for (int j = 0; j < n + 1; j ++ )
            scanf("%lf", &a[i][j]);

    int t = gauss();
    if (t == 2) puts("No solution");
    else if (t == 1) puts("Infinite group solutions");
    else
    {
        for (int i = 0; i < n; i ++ )
            printf("%.2lf\n", a[i][n]);
    }

    return 0;
}

#include <iostream>
#include <algorithm>

using namespace std;

const int N = 110;


int n;
int a[N][N];


int gauss()
{
    int c, r;
    for (c = 0, r = 0; c < n; c ++ )
    {
        int t = r;
        for (int i = r; i < n; i ++ )
            if (a[i][c])
                t = i;

        if (!a[t][c]) continue;

        for (int i = c; i <= n; i ++ ) swap(a[r][i], a[t][i]);
        for (int i = r + 1; i < n; i ++ )
            if (a[i][c])
                for (int j = n; j >= c; j -- )
                    a[i][j] ^= a[r][j];

        r ++ ;
    }

    if (r < n)
    {
        for (int i = r; i < n; i ++ )
            if (a[i][n])
                return 2;
        return 1;
    }

    for (int i = n - 1; i >= 0; i -- )
        for (int j = i + 1; j < n; j ++ )
            a[i][n] ^= a[i][j] * a[j][n];

    return 0;
}


int main()
{
    cin >> n;

    for (int i = 0; i < n; i ++ )
        for (int j = 0; j < n + 1; j ++ )
            cin >> a[i][j];

    int t = gauss();

    if (t == 0)
    {
        for (int i = 0; i < n; i ++ ) cout << a[i][n] << endl;
    }
    else if (t == 1) puts("Multiple sets of solutions");
    else puts("No solution");

    return 0;
}

其实就是层序遍历，我们用dfs来实现

1.先扩展根节点
2.再依次扩展左右儿子，也就是左右扩展下一层
3.再依次扩展下一层

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<int> printFromTopToBottom(TreeNode* root) {
        vector<int>res;
        if(!root) return res;
        queue<TreeNode*>q;
        q.push(root);

        while(q.size())
        {
            auto t=q.front();
            q.pop();
            res.push_back(t->val);
            if(t->left) q.push(t->left);
            if(t->right) q.push(t->right);
        }
        return res;
    }
};