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    中缀表达式求值 （+-*/^)

    #include <iostream>
#include <stack>
#include <string>
#include <map>
#include <cmath>
#define int long long

using namespace std;

stack <char> fu;
stack <int> num;
string st;
map <char, int> mp {{'+', 1}, {'-', 1}, {'*', 2}, {'/', 2}, {'^', 3}};

void init (string str) {
	for (int i = 0; i < str.size (); i ++) {
		if ((str[i] == '-' || str[i] == '+') && (i == 0 || str[i - 1] == '(')) {
			st += '0';
		}
		if (str[i] == '(' && str[i - 1] == ')') {
			st += '*';
		}
		st += str[i];
	}
}

void sum () {
	int b = num.top(); num.pop();
	int a = num.top(); num.pop();
	char c = fu.top(); fu.pop();

	int x;
	switch (c) {
		case '+' : num.push (a + b); break;
		case '-' : num.push (a - b); break;
		case '*' : num.push (a * b); break;
		case '/' : num.push (a / b); break;
		case '^' : num.push (pow (a, b)); break;
		default : break;
	}
	return ;
}

signed main() {
	string str;
	cin >> str;
	init (str);
	//cout << st << endl << endl;

	for (int i = 0; i < st.size (); i ++) {
		if (st[i] >= '0' && st[i] <= '9') {
			int j = i, x = 0;
			while (st[j] >= '0' && st[j] <= '9') //多位数;
				x = x * 10 + (st[j ++ ] - '0');
			num.push (x);
			i = j - 1;
		} 
	
		else if (st[i] == '(') {
			fu.push (st[i]);
		} 
	
		else if (st[i] == ')') {
			while (fu.top() != '(') { //算括号内的;
				sum ();
			}
			fu.pop (); //弹出 "(";
		} 
	
		else {
			while (fu.size() && fu.top() != '(' && mp[fu.top ()] >= mp[st[i]]) {
				sum ();
			} //栈不空 且 下一个符号不是左括号 且 当前优先级比栈顶优先级低;
			fu.push (st[i]);
		}
	}

	while (fu.size()) {
		sum ();
	}
	cout << num.top() << endl;
	return 0;
}

    dijkstra求最短路

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

using namespace std;

const int N = 510;

int n, m;
int g[N][N], dis[N], st[N];

int dijkstra () {
	memset (dis, 0x3f, sizeof dis);
	dis[1] = 0;

	for (int i = 1; i <= n; i ++) {
		int t = -1;
		for (int j = 1; j <= n; j ++) {
			if (st[j] == 0 && (t == -1 || dis[t] > dis[j])) {
				t = j;
			}
		}
	
		st[t] = true;
	
		for (int j = 1; j <= n; j ++) {
			dis[j] = min (dis[j], dis[t] + g[t][j]);
		}
	}

	if (dis[n] == 0x3f3f3f3f) {
		return -1;
	}
	return dis[n];
}

int main () {
	cin >> n >> m;
	memset (g, 0x3f, sizeof g);

	while (m --) {
		int a, b, c;
		cin >> a >> b >> c;
		g[a][b] = min (g[a][b], c); //判断重边； 
	}

	int t = dijkstra ();

	cout << t << endl;
	return 0;
}

    加一个堆优化

    #include <iostream>
#include <cstring>
#include <algorithm>
#include <queue>

using namespace std;

const int N = 1000010;
typedef pair <int, int> PII;

int n, m;
int h[N], w[N], e[N], ne[N]; 
int dis[N], st[N];
int idx;

void add (int a, int b, int c) {
	e[idx] = b, w[idx] = c, ne[idx] = h[a], h[a] = idx ++;
}

int dijkstra () {
	memset (dis, 0x3f, sizeof dis);
	dis[1] = 0;

	priority_queue <PII, vector <PII>, greater <PII> > p;
	p.push ({0, 1});

	while (!p.empty ()) {
		auto t = p.top ();
		p.pop ();
	
		int ver = t.second, dit = t.first;
	
		if (st[ver]) {
			continue;
		}
	
		st[ver] = true;
		for (int i = h[ver]; i != -1; i = ne[i]) {
			int j = e[i];
			if (dis[j] > dit + w[i]) {
				dis[j] = dit + w[i];
				p.push ({dis[j], j});
			}
		}
	}

	if (dis[n] == 0x3f3f3f3f) {
		return -1;
	}
	return dis[n];
}

int main () {
	cin >> n >> m;
	memset (h, -1, sizeof h);

	while (m --) {
		int a, b, c;
		cin >> a >> b >> c;
		add (a, b, c);
	}

	int t = dijkstra ();

	cout << t << endl;
	return 0;
}

    bellman-ford求最短路

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

using namespace std;

const int N = 510, M = 10010;
int n, m, k;
int dis[N], ba[N];

struct no {
	int a, b, w;
}ed[M];

int bellman_ford () {
	memset (dis, 0x3f, sizeof dis);
	dis[1] = 0;

	for (int i = 1; i <= k; i ++) {
		memcpy (ba, dis, sizeof dis);
	
		for (int j = 1; j <= m; j ++) {
			int a = ed[j].a, b = ed[j].b, w = ed[j].w;
			dis[b] = min (dis[b], ba[a] + w);
		}
	}

	if (dis[n] > 0x3f3f3f3f / 2) return -100000;
	return dis[n];
}

int main () {
	cin >> n >> m >> k;
	for (int i = 1; i <= m; i ++) {
		int a, b, w;
		cin >> a >> b >> w;
		ed[i] = {a, b, w};
	}

	int t = bellman_ford ();

	if (t == -100000) cout << "impossible\n";
	else cout << t << endl;
	return 0;
}

    SPFA求最短路

    #include <iostream>
#include <cstring>
#include <algorithm>
#include <queue>

using namespace std;

const int N = 1000010;
typedef pair <int, int> PII;

int n, m;
int h[N], w[N], e[N], ne[N]; 
int dis[N], st[N];
int idx;

void add (int a, int b, int c) {
	e[idx] = b, w[idx] = c, ne[idx] = h[a], h[a] = idx ++;
}

int spfa () {
	queue <int> q;
	memset (dis, 0x3f, sizeof dis);
	dis[1] = 0;

	q.push (1);
	st[1] = 1; //当时这个点在队列中 

	while (!q.empty ()) {
		int t = q.front ();
		q.pop ();
	
		st[t] = false;
	
		for (int i = h[t]; i != -1; i = ne[i]) {
			int j = e[i];
			if (dis[j] > dis[t] + w[i]) {
				dis[j] = dis[t] + w[i];
				if (!st[j]) {
					q.push (j);
					st[j] = 1;
				}
			}
		}
	}

	if (dis[n] == 0x3f3f3f3f) return -100000;
	else return dis[n];
}

int main () {
	cin >> n >> m;
	memset (h, -1, sizeof h);

	while (m --) {
		int a, b, c;
		cin >> a >> b >> c;
		add (a, b, c);
	}

	int t = spfa ();

	if (t == -100000) cout << "impossible\n";
	else cout << t << endl;
	return 0;
}

    SPFA判断负环

    #include <iostream>
#include <cstring>
#include <algorithm>
#include <queue>

using namespace std;

const int N = 1000010;
typedef pair <int, int> PII;

int n, m;
int h[N], w[N], e[N], ne[N]; 
int dis[N], st[N], cn[N];
int idx;

void add (int a, int b, int c) {
	e[idx] = b, w[idx] = c, ne[idx] = h[a], h[a] = idx ++;
}

int spfa () {
	queue <int> q;

	for (int i = 1; i <= n; i ++) {
		st[i] = 1;
		q.push (i);
	} 

	while (!q.empty ()) {
		int t = q.front ();
		q.pop ();
	
		st[t] = false;
	
		for (int i = h[t]; i != -1; i = ne[i]) {
			int j = e[i];
			if (dis[j] > dis[t] + w[i]) {
				dis[j] = dis[t] + w[i];
				cn[j] = cn[t] + 1;
			
				if (cn[j] > n) return 1;
				if (!st[j]) {
					q.push (j);
					st[j] = 1;
				}
			}
		}
	}

	return 0;
}

int main () {
	cin >> n >> m;
	memset (h, -1, sizeof h);

	while (m --) {
		int a, b, c;
		cin >> a >> b >> c;
		add (a, b, c);
	}

	if (spfa ()) cout << "Yes\n";
	else cout << "No\n";
	return 0;
}

    欧拉筛素数模板

    #include <iostream>
#include <cstring>

using namespace std;

const int N = 100000100;
const int n = 10000010;
int pr[N], f[N], cnt = 0;
//f[i] 为 1 表示是素数，否则为非素数。
//pr[i] 用来存素数。
 
void euler (int n) { //筛到n。 
	memset (f, 1, sizeof f); //开始默认全为素数。 
	f[1] = 0; //1不是素数。 

	for (int i = 2; i <= n; i ++) {
		if (f[i]) { //没有筛掉： 
			pr[++ cnt] = i; //i成为下一个素数。 
		}
	
		for (int j = 1; j <= cnt && pr[j] * i <= n; j ++) {
			//从pr[1]开始，逐个枚举已知的素数。
			//期望pr[j] 为 (i * pr[j]) 的最小质因子。
			//当然，i肯定比pr[j]大，因为pr[j]是在i之前得出的。 
		
			f[i * pr[j]] = 0; //标记为非素数。 
			if (i % pr[j] == 0) { //i中含有pr[j]这个因子： 
				break; //退出，保证线性复杂度。 
			}
		}
	}
}
int main () {
	ios :: sync_with_stdio (0);

	int q;
	cin >> q;
	euler(n);

	while (q --) {
		int k;
		cin >> k;
		cout << pr[k] << endl;
	}
	return 0;
}

    Copy

    DFS框架

    void dfs (int k) {
	if (/*所有空都已填完*/) {
		//判断最优解、记录答案
		return ; 
	}
	for (/*枚举这个空能填的选项*/) {
		if (/*这个选项是合法的*/) {
			//记录下这个空（保存现场）
			dfs (k + 1);
			//取消这个空（恢复现场） 
		}
	}
}

//********************************************************************

void dfs (int k) {
	for (/*枚举这个空能填的选项*/) {
		if (/*满足条件*/) {
			//保存结果
			if (/*到目的地*/) {
				//输出解 
			}  else {
				dfs (k + 1);
			} 
			//回复结果（回溯） 
		}
	}
}

    Copy

    BFS框架

    void bfs()
{
	//初始化，初始状态存入队列；
	//队列首指针head=0,尾指针tail=1;
	while(head<=tail)
	{
		//指针head后移一位，指向待扩展节点 
		for(int i=1;i<=max;i++)    //max为产生子节点的规则数
		{
			if(/*子节点符合条件*/)
			{
				//tail++,把新节点存入队列
				if(/*新节点与原条件重复*/)
				{
					//删除该节点（取消入队，tail--;）
					else if(/*新节点是目标节点*/) 
						//输出并退出
				}
			}
		} 
	} 
}

    Copy

    快速幂

    #include <bits/stdc++.h>
#define int long long
using namespace std;
int ksm (int a, int b, int p) {
	int res = 1;
	while (b) {
		if (b & 1) 
			res = res * a % p;
		b >>= 1;
		a = a * a % p;
	}
}
main () {
	int a, b, p;
	cin >> a >> b >> p;
	cout << ksm (a, b, p) << endl;
	return 0;
}

    Copy

    快速排序

    #include <bits/stdc++.h>
using namespace std;

const int N = 100010;
int n, q[N];

void qsort (int q[], int l, int r) {
	if (l >= r) return ;
	int x = q[l], i = l - 1, j = r + 1;
	while (i < j) {
		do i ++; while (q[i] < x);
		do j --; while (q[j] > x);
		if (i < j) swap (q[i], q[j]);
	}
	qsort (q, l, j);
	qsort (q, j + 1, r);
}
int main () {
	cin >> n;
	for (int i = 1; i <= n; i ++) 
		cin >> q[i];
	qsort (q, 1, n);
	for (int i = 1; i <= n; i ++)
		cout << q[i] << ' ';
	return 0;
}

    Copy

    归并排序

    #include <bits/stdc++.h>
using namespace std;

const int N = 1000010;
int n, q[N], tmp[N];

void msort (int q[], int l, int r) {
	if (l >= r) return ;
	int mid = l + r >> 1;
	msort (q, l ,mid);	msort (q, mid + 1, r);

	int k = 0, i = l, j = mid + 1;
	while (i <= mid && j <= r) {
		if (q[i] <= q[j]) tmp[k ++] = q[i ++];
		else tmp[k ++] = q[j ++];
	}

	while (i <= mid) tmp[k ++]= q[i ++];
	while (j <= r) tmp[k ++]= q[j ++];

	for (i = l, j = 0; i <= r; i ++, j ++)
		q[i] = tmp[j];
}
main () {
	cin >> n;
	for (int i = 1; i <= n; i ++) cin >> q[i];
	msort (q, 1, n);
	for (int i = 1; i <= n; i ++) cout << q[i] << ' ';
	return 0;
}

    Copy

    二分查找数的范围

    #include <bits/stdc++.h>
using namespace std;

const int N = 100010;
int n, m, q[N];
int main () {
	cin >> n >> m;
	for (int i = 0; i < n; i ++)
		cin >> q[i];
	while (m --) {
		int x;
		cin >> x;
	
		int l = 0, r = n - 1;
		while (l < r) {
			int mid = l + r >> 1;
			if (q[mid] >= x) r = mid;
			else l = mid + 1;
		}
		if (q[l] != x) cout << "-1 -1" << endl;
		else {
			cout << l << ' ';
		
			int l = 0, r = n - 1;
			while (l < r) {
				int mid = l + r + 1 >> 1;
				if (q[mid] <= x) l = mid;
				else r = mid - 1;
			}
		
			cout << l << endl;
		}
	}
	return 0;
}

    Copy

    #include <iostream>
#include <algorithm>

using namespace std;

int dep[1000000], cx[1000000];
struct no {
	int a;
}a[1000000];
int n, c;

bool cmp (no a, no b) {
	if (cx[a.a] == cx [b.a]) 
		return dep[a.a] < dep[b.a];
	else 
		return cx[a.a] > cx[b.a];
}
int main () {
	cin >> n >> c;
	for (int i = 1; i <= n; i ++) {
		cin >> a[i].a;
		if (dep[a[i].a] == 0) {
			dep[a[i].a] = i;
		}
		cx [a[i].a] ++;
	}

	sort (a + 1, a + 1 + n, cmp);

	for (int i = 1; i <= n; i ++) {
		cout << a[i].a << ' ';
	}
	return 0;
}

    Copy

    二分图的最大匹配

    #include <bits/stdc++.h>
using namespace std;
void Ios () {
	ios :: sync_with_stdio (false);
	cin.tie ();
	cout.tie ();
} 

const int N = 510, M = 100010;
int n1, n2, m;
int h[N], e[M], ne[M], idx;
int mac[N];
bool st[N];

int add (int a, int b) {
	e[idx] = b, ne[idx] = h[a], h[a] = idx ++;
}
bool find (int x) {
	for (int i = h[x]; i != -1; i = ne[i]) {
		int j = e[i];
		if (! st[j]) {
			st[j] = true;
			if (mac[j] == 0 || 	find (mac[j])) {
				mac[j] = x;
				return true;
			}
		}
	}
	return false;
}
int main () {
	Ios ();
	cin >> n1 >> n2 >> m;
	memset (h, -1, sizeof h);

	while (m --) {
		int a, b;
		cin >> a >> b;
		add (a, b);
	}

	int res = 0;
	for (int i = 1; i <= n1; i ++) {
		memset (st, false, sizeof st);
		if (find (i)) res ++;
	}
	cout << res << endl;
	return 0;
}

    Copy

    高精度加法

    #include <bits/stdc++.h>
using namespace std;
vector<int> add (vector<int> &A, vector<int> &B) {
    if (A.size () < B.size ()) 
		return add(B, A);

    vector <int> C;
    int t = 0;
    for (int i = 0; i < A.size (); i ++ ) {
        t += A[i];
        if (i < B.size()) t += B[i];
        C.push_back(t % 10);
        t /= 10;
    }

    if (t) C.push_back(t);
    return C;
}

int main () {
    string a, b;
    vector<int> A, B;
    cin >> a >> b;
    for (int i = a.size() - 1; i >= 0; i -- ) A.push_back(a[i] - '0');
    for (int i = b.size() - 1; i >= 0; i -- ) B.push_back(b[i] - '0');
    auto C = add(A, B);
    for (int i = C.size() - 1; i >= 0; i -- ) cout << C[i];
    cout << endl;
    return 0;
}

    Copy

    高精度减法

    #include <bits/stdc++.h>
using namespace std;

bool cmp (vector <int> &a, vector <int> &b) {
	if (a.size () != b.size ()) {
		return a.size () > b.size ();
	}
	for (int i = a.size (); i >= 0; i --)
		if (a[i] != b[i])
			return a[i] > b[i];
	return true;
}
vector<int> sub (vector<int> &a, vector<int> &b) {
    vector <int> c;
	for (int i = 0, t = 0; i < a.size (); i ++) {
		t = a[i] - t;
		if (i < b.size ()) t -= b[i];
		c.push_back ((t + 10) % 10);
		if (t < 0) t = 1;
		else t = 0;
	}

	while (c.size () > 1 && c.back () == 0)
		c.pop_back ();
	
	return c;
}

int main () {
    string a, b;
    vector<int> A, B;
    cin >> a >> b;
  
    for (int i = a.size() - 1; i >= 0; i -- ) A.push_back (a[i] - '0');
    for (int i = b.size() - 1; i >= 0; i -- ) B.push_back (b[i] - '0');
  
    if (cmp (A, B)) {
    	auto C = sub (A, B);
    	for (int i = C.size() - 1; i >= 0; i -- ) cout << C[i];
	} else {
		auto C = sub (B, A);
		cout << "-";
		for (int i = C.size() - 1; i >= 0; i -- ) cout << C[i];
	}
    cout << endl;
    return 0;
}

    Copy

    高精度乘法（A*b）

    #include <bits/stdc++.h>

using namespace std;

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;
    }

    while (C.size() > 1 && C.back() == 0)
		C.pop_back();

    return C;
}

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

    vector <int> A;
    for (int i = a.size() - 1; i >= 0; i --) A.push_back(a[i] - '0');

    auto C = mul(A, b);

    for (int i = C.size() - 1; i >= 0; i --) {
        cout << C[i];
    }

    return 0;
}

    Copy

    高精度除法

    #include <bits/stdc++.h>

using namespace std;

vector <int> div(vector <int> & A, int b, int &r) {
    vector <int> C;
	r = 0;

    for (int i = A.size () - 1; i >= 0; i --) {
        r = r * 10 + A[i];
        C.push_back (r / b);
        r %= b;
    }
  
    reverse (C.begin(), C.end ());
    while (C.size()>1&&C.back()==0) 
		C.pop_back();

    return C;
}

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

    vector <int> A;
    for (int i = a.size() - 1; i >= 0; i --) A.push_back(a[i] - '0');

	int r;
    auto C = div(A, b, r);

    for (int i = C.size() - 1; i >= 0; i --) {
        cout << C[i];
    }
	cout << endl << r << endl; 
    return 0;
}

    #include <iostream>
#include <map>

using namespace std;

int n, m, ans = 0, a[2000010], z[2000010];
map <int, int> mp;

int main () {
	cin >> n >> m;
	for (int i = 1; i <= n; i ++) {
		cin >> a[i];
		z[i] = z[i - 1] + a[i];
		mp[z[i - 1]] ++;
		ans += mp[z[i] - m];
	}
	cout << ans << endl;
	return 0;
}

    Copy

    AC自动机

    #include <bits/stdc++.h>
using namespace std;

const int N = 10010, S = 55, M = 1000010;
int n;
int tr[N * S][26];
int cnt[N * S];
int idx;
char str[M];
int q[N * S];
int ne[N * S];

void insert() {

    int p = 0;
    for (int i = 0; str[i]; i++) {
        int t = str[i] - 'a';
        if (!tr[p][t]) tr[p][t] = ++idx;
        p = tr[p][t];
    }
    cnt[p]++;
}

void build() {

    int hh = 0, tt = -1;
  
    for (int i = 0; i < 26; i++)
        if (tr[0][i])
            q[++tt] = tr[0][i];

    while (hh <= tt) {
        int t = q[hh++];
        for (int i = 0; i < 26; i++) {
            int c = tr[t][i];
            if (!c) continue;

            int j = ne[t];
            while (j && !tr[j][i]) j = ne[j];
            if (tr[j][i]) j = tr[j][i];
            ne[c] = j;
            q[++tt] = c;
        }
    }
}

int main() {

    int T;
    cin >> T;
    while (T--) {
        memset(tr, 0, sizeof tr);
        memset(cnt, 0, sizeof cnt);
        memset(ne, 0, sizeof ne);
        idx = 0;

        cin >> n;
        for (int i = 0; i < n; i++) {
            scanf("%s", str);
            insert();
        }

        build();

        scanf("%s", str);
        int res = 0;
        for (int i = 0, j = 0; str[i]; i++) {
            int t = str[i] - 'a';
            while (j && !tr[j][t]) j = ne[j];
            if (tr[j][t]) j = tr[j][t];

            int p = j;
            while (p) {
                res += cnt[p];
                cnt[p] = 0;
                p = ne[p];
            }
        }

        cout << res << endl;
    }

    return 0;
}

    Copy

    树的节点计数

    #include <iostream>

using namespace std;

const int N = 50010;
int n, x, y;
int now[2 * N], pre[2 * N], son[2 * N], tot, deep[N];

void put (int x, int y) {
	pre[++ tot] = now[x];
	now[x] = tot;
	son[tot] = y;
}

void dfs (int dep, int fa) {
	deep[dep] = 1;
	for (int i = now[dep]; i; i = pre[i]) {
		if (son[i] != fa) {
			dfs (son[i], dep);
			deep[dep] += deep[son[i]];
		}
	}
}

int main () {
	cin >> n;
	for (int i = 1; i < n; i ++) {
		cin >> x >> y;
	
		put (x, y);
		put (y, x);
	}
	dfs (1, 0);
	for (int i = 1; i <= n; i ++) {
		cout << deep[i] - 1<< endl; 
	}
}

    Copy

    cout<<fixed<<setprecision(n)<<a;

    Sublime Text ：

    1.打开 SublimeSublime TextText 软件;

    2.打开 ToolsTools => BuildBuild SystemSystem => NewNew BuildBuild SystemSystem ;

    3.将源代码改为：

    {
    "shell_cmd": "g++ \"${file}\" -o \"${file_path}/${file_base_name}\" -Wall -std=c++11 -Wextra -g",
    "file_regex": "^(..[^:]*):([0-9]+):?([0-9]+)?:? (.*)$",
    "working_dir": "${file_path}",
    "selector": "source.c, source.c++",
 
    "variants":
    [
        {
            "name": "Run",
            "shell_cmd" : "start cmd /c \"\"${file_path}/${file_base_name}\" & pause\""
        },
        {
            "name": "RunInCommand",
            "shell_cmd" : "g++ \"${file}\" -o \"${file_path}/${file_base_name}\" -Wall -std=c++11 -Wextra -g && start cmd /c \"\"${file_path}/${file_base_name}\" & echo.                                & echo -------------------------------- & echo Process exited after seconds with return value 0 & pause\""
        },
        {
            "name": "Debug",
            "shell_cmd" : "start cmd /c \"gdb \"${file_path}/${file_base_name}\"\""
        }
    ]
}

    4.将文件名改为：aa.sublime-build;

    5.打开 PreferencesPreferences => KeyKey BindingsBindings，进入键位设置；

    6.按下 Ctrl+F 快捷键，在搜索栏输入 f7；

    7.将高亮代码 f7 下方的 Ctrl+B 改为 F11，并保存；

    8.最后，新建一个源代码，将其命名为 xx.cpp，就能编译了。。。

    用位运算实现加减乘除：

    1. 加法：

    // 递归写法
int add(int num1, int num2) {
	if (num2 == 0)
		return num1;
	int sum = num1 ^ num2;
	int carry = (num1 & num2) << 1;
	return add(sum, carry);
}

// 迭代写法
int add(int num1, int num2) {
	int sum = num1 ^ num2;
	int carry = (num1 & num2) << 1;
	while (carry != 0) {
		int a = sum;
		int b = carry;
		sum = a ^ b;
		carry = (a & b) << 1;
	}
	return sum;
}

    扩展欧几里得算法求ax+by=c：

    void ojld (int a, int b) {
        if (b == 0) {
		x = 1, y = 0;
		return ;
	}
	ojld (b, a % b);

	t1 = x, t2 = y;
	x = t2, y = t1 - (a / b) * t2;
}

    快速幂：

    int qmi (int a, int b, int p) {
    int ans = 1;
	while (b) {
		if (b & 1) {
			ans = (a * a) % p;
			b --;
		}
		b /= 2;
		a = a * a % p;
     }
     return ans;
}

    三种筛素数的方法：

      1. SBSB筛法：
    if (n <= 1) return 0;
for (int i = 2; i * i <= n; i ++)
	if (n % i == 0)
		return 0;
return 1;
      1. 埃式筛法：
    int p[1000010];
for (int i = 2; i <= sqrt (n); i ++) {
	if (p[i] == 0) {
		for (int j = 2; j <= n / i; j ++) {
			p[i * j] = 1;
		}
	}
}
if (p[n] == 1) return 0;
else return 1;
      1. 欧拉筛法：
    // 筛 1 ~ N 中的素数
ok[1] = 1;
for (int i = 2; i <= N; i ++) {
    if (!ok[i]) {
        pr[++ t] = i;
    }
    for (int j = 1; j <= t && i * pr[j] <= N; j ++) {
        ok[i * pr[j]] = 1;
        if (i % pr[j] == 0) {
            break;
        }
    }
    continue;
}
for (int i = 1; i <= 100; i ++) {
    if (!ok[i]) {
        cout << i << ' ';
    }
}

    // 筛 1 ~ N 中与N互质的数
void euler2 () {
	phi[1] = 1;
	for (int i = 2; i <= N; i++) {
		if (!isprime[i]) {
			prime[++tot] = i;
			phi[i] = i - 1;
		}
		for (int j = 1; j <= tot; j++) {
			if (i * prime[j] > N)
				break;
			isprime[i * prime[j]] = 1;
			if (i % prime[j] == 0) {
				phi[i * prime[j]] = phi[i] * prime[j];
				break;
			} else
				phi[i * prime[j]] = phi[i] * (prime[j] - 1);
		}
	}

	for (int i = 1; i <= N; i ++) {
		cout << phi[i] << ' ';
	}

}

    准备工作

    1. 比赛前一天晚上请准备好你的各种证件以及笔等基本工具（我的身边就有因为没带准考证而爆炸的例子）。
    2. CSP 是允许带食品和饮料的，所以可以带一些水和巧克力等进入考场补充体力。
    3. 比赛开始前请先调整你的屏幕分辨率到你喜欢的大小。
    4. 比赛开始前请把编译器的字体调为你平时惯用的字体。
    5. 在不影响视野的情况下，请将字号尽可能调大，方便查错（特别是一些毒瘤模拟或搜索）
    6. 比赛开始前多敲敲键盘，顺手的键盘都需要敲打一番（提前熟悉键盘的各个键——经常会与自己学校的不一样，在自己学校比赛的请忽略这句话），这段时间里你可以先打好快读快写、头文件等模板。
    7. 根据个人需求并在合理时间内待在考场里，熟悉一下气氛。

    参考文献 https://www.scratch5.com/scratch#621914581663660938942 https://www.acwing.com/blog/content/829/ https://www.acwing.com/blog/content/798/ https://www.cnblogs.com/ljy-endl/p/11768512.html https://www.cnblogs.com/dysyn1314/p/13934854.html https://www.cnblogs.com/codingxu/p/15433809.html https://www.cnblogs.com/milkycoffee/p/csp-s-2021-zhuyi.html https://www.cnblogs.com/skip1978/p/11861737.html

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