【PAT笔记】常用子函数

1.gcd（最大公因数）

int gcd(int a,int b) {
	if (b == 0)return a;
	else return gcd(b, a%b);
}

2.判断素数（不是素数筛）

bool prime(ll it) {
	if (it < 2)return false;
	for (ll i = 2; i*i <= it; i++)
		if (it%i == 0)return false;
	return true;
}

3.素数筛

   （1）埃氏筛

void Eratosthenes(int it) {
	int cnt = 0, int prime[10000], f[10000];
	for (int i = 2; i*i <= it; i++)
		if (!f[i])for (int j = i * 2; j <= it; j += i)f[j] = 1;
	for (int i = 2; i <= it; i++)
		if (!f[i])prime[cnt++] = i;
}​

（2）欧拉筛

void Euler(int n) {
	int cnt = 0, prime[100000], f[100000];
	for (int i = 2; i <= n; i++) {
		if (!f[i]) 	prime[cnt++] = i;
		for (int j = 0; j < cnt; j++) {
			if (prime[j] * i > n)break;
			f[prime[j] * i] = 1;
			if (i%prime[j] == 0)break;
		}
	}
}

4.高精度（java BigInteger）

import java.math.BigInteger;
import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner scanner = new Scanner(System.in);
		BigInteger bigInteger = scanner.nextBigInteger();
		BigInteger bigInteger2 = scanner.nextBigInteger();
		System.out.println(bigInteger.add(bigInteger2));
	}
}

 5.并查集

int tree[1005];

void union_find(int a,int b) {
	if (a == b)return;
	else if (a < b)tree[b] = a;
	else tree[a] = b;
}
int find_root(int it) {
	if (tree[it] == -1)return it;
	else return tree[it] = find_root(tree[it]);
}

int main() {
	memset(tree, -1, 1000);
	union_find(find_root(10), find_root(5));
	return 0;
}

6.dijkstra

void dijkstra(int s,int e) {
	dis[s] = 0;		//起点距离赋值为0
	int it = s;				//迭代器
	while (!f[it]) {
		f[it] = 1;			//标记it的信息已经用过了
		for(int i=0;i<100;i++)		//用it的信息更像图像距离
			if (dis[it] + graph[it][i] < dis[i]) {
				dis[i] = dis[it] + graph[it][i];
				pre.clear();		//如果找到更近的点，此前的需要清空
				pre[i].push_back(it);
			}
			else if (dis[it] + graph[it][i] == dis[i]) {
				pre[i].push_back(it);	//距离一样的点，加入i的前驱集合
			}
			int minl = INF, next = it;    //开始寻找下一个点
			for(int i=0;i<100;i++)
				if (!f[i] && dis[i] < minl) {
					minl = dis[i];
					next = i;
				}
			it = next;
	}
}

7.DFS路径回溯

 

int graph[100][100],f[100];
vectorpath, ans;
void dfs(int s,int e) {
	if (s == e) {
		ans = path;	
		return;
	}
	for(int i=0;i<100;i++)
		if (!f[i]) {
			f[i] = 1;
			path.push_back(i);
			dfs(i, e);
			f[i] = 0;
			path.pop_back();
		}
}

8.Manacher

int Mannncher(char s[],int length) {
	int id, mx = 0, len = 1, p[100] = { 0 };	//mx：最大回文串的右边界   id：回文串的中心
	string str="\n";		//这里以回车符为填充符号
	for (int i = 0; i < length; i++) {
		str.push_back(s[i]);
		str.push_back('\n');
	}
	for (int i = 1; i < str.size(); i++) {
		if (i < mx)p[i] = min(p[id*2-i],mx-id);	
		//第一个if为第一类情况，大串套小串，而且对称位置的小串已经算出来了。
		else p[i] = 1;
		//不然就是初始为1赋值
		while (str[i - p[i]] == str[i + p[i]])++p[i];
		if (p[id] < p[i]) {		//以str[i]为中心向两边拓展
			id = i;
			mx = p[id] + i;
		}
		len = max(len,p[i]-1);//更新最大值
	}
}

9.后中建树（前中同理）

int poster[100], inorder[100];
struct node {
	int v;
	node* left, *right;
};
node *tree;

void build(int inl,int inr,int posl,int posr,node* &tree) {
	if (inl > inr)return;
	int v = poster[posr],it = inl;
	for (; it <= inr&&inorder[it] != v; it++);
	tree = (node*)malloc(sizeof(node));
	tree->v = v;
	tree->left = tree->right = NULL;
	build(inl,it-1,posl,posl+it-inl-1,tree->left);
	build(it + 1, inr, posl + it - inl, posr - 1, tree->right);
}