无风扇嵌入式车载电脑在矿山车辆行业应用

1.基础算法

（1）一维前缀和

#include<iostream>

using namespace std;

const int N = 1e5+10;

int p[N],res[N];
int n,Q,l,r;
 
int main()
{
	cin >> n >> Q;
	
	for(int i = 1;i<=n;++i)
	{
		cin >> p[i];
		res[i] = res[i - 1] + p[i];
	}
	
	while(Q--){
		cin >> l >> r;
		cout << res[r] - res[l - 1] << endl;
	}
	
	return 0;
	
}

（2）二维前缀和

#include<iostream>

using namespace std;

const int N = 1005;

int p[N][N],res[N][N];
int n,m,Q,x1,x2,y1,y2;

int main()
{
	cin >> n >> m >> Q;
	
	for(int i = 1;i <= n; ++i){
		for(int j = 1;j<=m;++j){
			cin >> p[i][j];
			res[i][j] = res[i - 1][j] + res[i][j - 1] - res[i - 1][j - 1] + p[i][j];
		}
	} 
	
	while(Q--)
	{
		cin >> x1 >> y1 >> x2 >> y2;
		
		cout << res[x2][y2] + res[x1 - 1][y1 - 1] - res[x1 - 1][y2] - res[x2][y1 - 1] << endl; 
	}
	
	return 0;
	
}

（3) 一维差分

#include<iostream>

using namespace std;

const int N = 1e5 + 10;

int p[N],res[N];
int n,Q,l,r,x;

int main(){
	cin >> n >> Q;
	
	for(int i = 1;i <= n;++i){
		cin >> p[i];
		res[i] = p[i] - p[i - 1];
	}
	
	while(Q--)
	{
		cin >> l >> r >> x;
		res[l] += x;
		res[r + 1] -= x;
	}
	
	int tot = 0;
	
	for(int i = 1;i<=n;++i)
	{
		tot += res[i];
		cout << tot << ' ';
		
	}
	
	return 0;
}

（4）二维差分

#include<iostream>

using namespace std;

const int N = 1005;

int p[N][N],res[N][N];

int n,m,Q,x1,y1,x2,y2,c;

void add(int x1,int y1,int x2,int y2,int c)
{
	res[x1][y1] += c;
	res[x2 + 1][y2 + 1] += c;
	res[x1][y2 + 1] -= c;
	res[x2 + 1][y1] -= c; 
}

int main()
{
	cin >> n >> m >> Q;
	for(int i = 1;i<=n;++i){
		for(int j = 1;j<=m;++j){
			cin >> p[i][j];
			add(i,j,i,j,p[i][j]);
		}
	}
	
	while(Q--)
	{
		cin >> x1 >> y1 >> x2 >> y2 >> c;
		add(x1,y1,x2,y2,c);
	}
	
	for(int i = 1;i <= n; ++i)
	{
		for(int j = 1;j <= m; ++j)
		{
			res[i][j] += res[i - 1][j] + res[i][j - 1] - res[i - 1][j - 1];
			cout << res[i][j] << ' ';
		}
		cout << endl;
	}
	
	return 0;
	
} 

（5）双指针

#include<iostream>

using namespace std;

const int N = 1e5 + 10;

int cnt[N],p[N];
int n,len = -1;

int main(){
	cin >> n;
	
	for(int i = 1;i<=n;++i){
		cin >> p[i];
	}
	
	for(int i = 1,j = 1;j <= n;++j){
		
		cnt[p[j]]++;
		
		while(cnt[p[j]] > 1)
		{
			cnt[p[i]]--;
			i++;
		}
		
		len = max(j - i + 1,len);	
	}
	
	cout << len << endl;
	
	return 0;
}

（6）区间合并

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

using namespace std;

const int N = 1e5 + 10;

typedef pair<int,int>PII;

vector<PII>q;

int n;
int cnt = 0;

int main(){
	cin >> n;
	
	for(int i = 1;i <= n; ++i){
		int l,r;
		cin >> l >> r;
		q.push_back({l,r}); 
	}
	
	sort(q.begin(),q.end());

	
	for(int i = 0;i < q.size();++i)
	{
		int l,r;
		if(i == 0)
		{
			cnt ++;
			l = q[i].first;
			r = q[i].second;	
			continue;
		}
		
		if(q[i].first <= r) r = max(q[i].second,r);
		else{
			cnt ++;
			l = q[i].first;
			r = q[i].second;
		}
		
	}
	
	cout << cnt << endl;
	return 0;
} 

（7）归并排序与逆序对

#include<iostream>
 #include<vector>
 
 
 using namespace std; 
 
 const int N = 1e5 + 10;
 
 int n,p[N];
 int cnt = 0;
 
 void merge(int l,int r)
 {
 	if(l >= r) return;
	 
	vector<int>t;
 	t.clear();
 	 	
 	int mid = (l + r) / 2;
 	merge(l,mid),merge(mid + 1,r);
 	
 	int i = l,j = mid + 1;
 	
 	while(i <= mid && j <= r)
	{
		if(p[i] > p[j])
		{
			cnt += mid - i + 1;
			t.push_back(p[j]);
			j++;
		}
		
		if(p[i] <= p[j])
		{
			t.push_back(p[i]);
			i++;
		}
	}
	
	while(i <= mid){
		t.push_back(p[i]);
		i++;
	}
	
	while(j <= r){
		t.push_back(p[j]);
		j++;
	}
 	
 	for(int i = 0;i < t.size(); ++i) p[i + l] = t[i];
	 
	return; 
 }
 
 int main()
 {
 	cin >> n;
 	
 	for(int i = 1;i <= n; ++i) cin >> p[i];
 	
 	merge(1,n);
 	
 	cout << cnt << endl;
 	
 	for(int i = 1;i <= n; ++i) cout << p[i] << ' ';
 	
 	return 0;
 	
 	
 }

（8）单调栈

#include<iostream>
#include<stack>

using namespace std;

int n,p;
stack<int>q;

int main()
{
	
	cin >> n;
	
	while(n--)
	{
		cin >> p;
		
		while(q.size() != 0 && q.top() >= p)
		{
			q.pop();
		}
		
		if(q.size() == 0) cout << -1 << ' ';
		else cout << q.top() << ' '; 
		
		q.push(p); 
	}
	
	return 0;
}

（9）单调队列

#include<iostream>

using namespace std;

const int N = 1e6 + 10;

int q[N],p[N];
int n,len;

int main()
{
	cin >> n >> len;
	
	for(int i = 0;i < n; ++i) cin >> p[i];
	
	int h,t;
	
	h = 0,t = -1;
	for(int i = 0;i < n; ++i)
	{
		while(h <= t && q[h] < i - len + 1) h++;
		while(h <= t && p[q[t]] >= p[i]) t--;
		t++;
		q[t] = i;
		if(i >= len - 1) cout << p[q[h]] << ' ';
	}
	
	cout << endl;
	
	h = 0,t = -1;
	for(int i = 0;i < n; ++i)
	{
		while(h <= t && q[h] < i - len + 1) h++;
		while(h <= t && p[q[t]] <= p[i]) t--;
		t++;
		q[t] = i;
		if(i >= len - 1) cout << p[q[h]] << ' ';
	}
	
	
	return 0;
} 

（10）整数二分

#include<iostream>

using namespace std;

const int N = 1e5 + 10;

int p[N];
int n,Q,x;

int leftfind(int x)
{
	int l = 1,r = n;
	
	while(l < r)
	{
		int mid = (l + r) / 2;
		if(p[mid] < x) l = mid + 1;
		else r = mid;
	}
	
	if(p[l] == x) return l - 1;
	else return -1;
}

int rightfind(int x){
	int l = 1,r = n;
	
	while(l < r){
		int mid = (l + r + 1) / 2;
		if(p[mid] > x) r = mid - 1;
		else l = mid;
	}
	
	if(p[l] == x) return l - 1;
	else return -1;
}

int main(){
	cin >> n >> Q;
	for(int i = 1;i <= n; ++i) cin >> p[i];
	
	while(Q--)
	{
		cin >> x;
		cout << leftfind(x) << ' ';
		cout << rightfind(x) << endl;	
	}
	
	return 0;
} 

2.数论

（1）快速幂

#include<iostream>

using namespace std;

typedef long long LL;

int Q;

LL qmi(LL a,LL b,LL mod){
	
	LL res = 1;
	while(b)
	{
		if(b&1) res = res * a % mod;
		a = a * a % mod;
		b = b >> 1;
	}
	
	return res;
}

int main()
{
	cin >> Q;
	
	while(Q--)
	{
		LL a,b,c;
		cin >> a >> b >> c;
		cout << qmi(a,b,c) << endl;
	}
	
	return 0;
} 

（2）约数个数

#include<unordered_map>
#include<iostream>

using namespace std;

unordered_map<int,int>q;
unordered_map<int,int>::iterator it;

const int mod = 1e9 + 7;

typedef long long LL; 

void handle(int x){
	
	for(int i = 2;i * i <= x;++i)
	{
		if(x % i == 0)
		{
			int cnt = 0;
			while(x % i == 0){
				x = x / i;
				cnt++;
			}
			q[i] += cnt;
		}
	}
	if(x > 1) q[x] += 1;
	
}

int Q,x;
LL ans = 1;

int main()
{
	cin >> Q;
	
	while(Q--)
	{
		cin >> x;
		handle(x); 
	}
	
	for(it = q.begin();it != q.end();++it){
		int a = it->first;
		int b = it->second;
		
		LL res = 1;
		
		for(int i = 1;i <= b; ++i)
		{
			LL res = (res * a + 1) % mod;
		}
		
		ans = ans * res % mod;
		
		
	}
	
	cout << ans << endl;
}

（3）约数之和

#include<unordered_map>
#include<iostream>

using namespace std;

unordered_map<int,int>q;
unordered_map<int,int>::iterator it;

const int mod = 1e9 + 7;

typedef long long LL; 

void handle(int x){
	
	for(int i = 2;i * i <= x;++i)
	{
		if(x % i == 0)
		{
			int cnt = 0;
			while(x % i == 0){
				x = x / i;
				cnt++;
			}
			q[i] += cnt;
		}
	}
	if(x > 1) q[x] += 1;
	
}

int Q,x;
LL ans = 1;

int main()
{
	cin >> Q;
	
	while(Q--)
	{
		cin >> x;
		handle(x); 
	}
	
	for(it = q.begin();it != q.end();++it)
	{
		int a = it->first;
		int b = it->second;
	
		LL res = 1;
		
		while(b--) 
		{
			res = (res * a + 1) % mod;
		}
		
		ans = ans * res % mod;
			
	}
	
	cout << ans << endl;
	
	return 0;
}

（4）组合数（一）

#include<iostream>

using namespace std;

typedef long long LL;

const int mod = 1e9 + 7;
const int N = 2005;

int c[N][N];
int main()
{
	
	for(int i = 0;i < N; ++i){
		for(int j = 0;j <= i; ++j)
		{
			if(j == 0) c[i][j] = 1;
			else c[i][j] = (c[i - 1][j - 1] + c[i - 1][j]) % mod;
		}
	}
	
	int Q;
	cin >> Q;
	
	while(Q--)
	{
		int a,b;
		cin >> a >> b;
		
		cout << c[a][b] << endl; 
	}
	
	return 0;
}

（5）组合数（二）

#include<iostream>

using namespace std;

const int N = 1e5 + 5;
const int mod = 1e9 + 7;

typedef long long LL;

LL f[N],inf[N];


LL qmi(LL a,LL b,LL mod)
{
	LL res = 1;
	
	while(b)
	{
		if(b & 1) res = res * a % mod;
		a = a * a % mod;
		b = b >> 1;
		
	}
	
	return res;
} 

int main()
{
	f[0] = inf[0] = 1;
	for(int i = 1;i < N;++i)
	{
		f[i] = f[i - 1] * i % mod;
		inf[i] = qmi(f[i],mod - 2,mod);
	}
	
	int Q,a,b;
	cin >> Q;
	while(Q--)
	{
		cin >> a >> b;
		
		LL ans = f[a] * inf[b] % mod * inf[a - b] % mod;
		
		cout << ans << endl;
	}
	
	return 0;	
	
} 

（6）位运算

#include<iostream>

using namespace std;

int main(){
	int Q,x,cnt = 0;
	
	cin >> Q;
	
	while(Q--)
	{
		cnt = 0;
		cin >> x;
	
	    while(x)
		{
	  	  if(x&1) cnt++;
		  x = x >> 1;
	    }
	
	    cout << cnt << ' ';
	}
	return 0;
}

（7）快速幂求解逆元

#include<iostream>

using namespace std;

typedef long long LL;

LL qmi(LL a,LL b,LL mod){
	LL res = 1;
	
	while(b){
		
		if(b & 1) res = res * a % mod;
		a = a * a % mod;
		b = b >> 1;
		
	}
	
	return res;
} 

int main(){
	int Q;
	cin >> Q;
	
	while(Q--){
		LL a,p;
		cin >> a >> p;
		if(a % p == 0){
			cout << "impossible" << endl;
			continue; 
		}
		
		cout << qmi(a,p - 2,p) << endl;
	}
	
	return 0;
}

（8）gcd

#include<iostream>

using namespace std;

int gcd(int a,int b)
{
	return b?gcd(b,a%b):a;
}

int Q,a,b;

int main()
{
	cin >> Q;
	
	while(Q--)
	{
		cin >> a >> b;	
		cout << gcd(a,b) << endl;
	}
	
	return 0;
	
	
}

（9）欧拉函数

#include<iostream>

using namespace std;

typedef long long LL;

LL euler(int x)
{
	LL res = x;
	
	for(int i = 2;i * i <= x;++i){
		if(x % i == 0)
		{
			res = res * (i - 1) / i;
			while(x % i == 0) x = x/i;
		}	
	}
	
	if(x > 1) res = res * (x - 1) / x; 
	
	return res;
}

int main(){
	int Q,x;
	
	cin >> Q; 
	
	while(Q--)
	{
		cin >> x;
		cout << euler(x) << endl;
	} 
	
	return 0;
}

3.图论

（1）djistra

#include<iostream>
#include<queue>
#include<cstring>
#include<vector>

using namespace std;

const int N = 1e6 + 10;

int dist[N];
bool st[N];
int ne[N],e[N],w[N],h[N],idx;

int n,m;

typedef pair<int,int>PII;
priority_queue<PII,vector<PII>,greater<PII>>q;

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

int djistra()
{
	dist[1] = 0;
	q.push({0,1});
	
	while(q.size() != 0)
	{
		PII t = q.top();
		q.pop();
		
		int v = t.second,d = t.first;
		
		if(st[v]) continue;
		st[v] = true;
		
		for(int i = h[v];i != -1;i = ne[i])
		{
			int j = e[i];
			if(dist[j] > d + w[i])
			{
				dist[j] = d + w[i];
				q.push({dist[j],j});
			}
			
		}	
	}  
	
	if(dist[n] == 0x3f3f3f3f) return -1;
	else return dist[n];
	
}

int main()
{
	memset(h, -1 ,sizeof(h));
	memset(dist, 0x3f, sizeof(dist));
	
	cin >> n >> m;
	
	for(int i = 1;i <= m; ++i)
	{
		
		int a,b,c;
		
		cin >> a >> b >> c;
		
		add(a,b,c);
	}
	
	cout << djistra() << endl;
	
	return 0;
	
}

（2）floyd

#include<iostream>
#include<cstring>

using namespace std;

const int N = 205;

int p[N][N];

int n,m,k;

void floyd(){
	
	for(int k = 1;k <= n; ++k){
		for(int i = 1;i <= n; ++i){
			for(int j = 1;j <= n; ++j){
				if(p[i][j] > p[i][k] + p[k][j]) p[i][j] = p[i][k] + p[k][j];
			}
		}
	}
}

int main()
{
	
	cin >> n >> m >> k;
	
	memset(p,0x3f,sizeof(p));
	
	for(int i = 1;i <= n; ++i) p[i][i] = 0;
 	
	for(int i = 1;i <= m; ++i)
	{
		int a,b,c;
		cin >> a >> b >> c;
		p[a][b] = min(p[a][b],c);
	}
	
	
	floyd(); 

	
	while(k--)
	{
		int a,b;
		
		cin >> a >> b;
		
		if(p[a][b] > 0x3f3f3f3f/2) cout << "impossible" << endl;
		else cout << p[a][b] << endl;
		
	}
	
	return 0;
	
	
}

（3）spfa

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

using namespace std;

const int N = 1e6 + 10;

int dist[N];
int h[N],e[N],ne[N],w[N],idx;
bool st[N];

int n,m;

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

void spfa()
{
	queue<int>q;
	
	dist[1] = 0;
	st[1] = true; 
	
	q.push(1);
	
	while(q.size() != 0)
	{
		auto t = q.front();
		q.pop();
		
		st[t] = false;
		
		for(int i = h[t]; i != -1; i = ne[i])
		{
			int j = e[i];
			
			if(dist[j] > dist[t] + w[i])
			{
				dist[j] = dist[t] + w[i];
				
				if(st[j]) continue;
				st[j] = true;
				q.push(j);		
			}	
		}	
	}
	
	if(dist[n] == 0x3f3f3f3f) cout << "impossible" << endl;
	else cout << dist[n] << endl;	
}


int main()
{
	memset(h,-1,sizeof(h));
	memset(dist,0x3f,sizeof(dist));
	
	cin >> n >> m;
	
	for(int i = 1;i <= m;++i)
	{
		int a,b,c;
		cin >> a >> b >> c;
		add(a,b,c);
	}
	
	spfa();
	
	return 0; 
}

（4）树的遍历

#include<iostream>
#include<cstring>

using namespace std;

const int N = 1e5 + 10;

bool st[N];
int e[N * 2],ne[N * 2],h[N * 2],idx = 0;
int n,ans = 0x3f3f3f3f;

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

int dfs(int u)
{
	st[u] = true;
	int sum = 0,tot = -1;
	for(int i = h[u];i != -1;i = ne[i]){
		
		int j = e[i];
		
		if(st[j])
		{
		  int k = dfs(j);
		  sum += k;
		  tot = max(tot,k);
			
		}		
		
	}
	
	tot = max(n - 1 - sum,tot);
	ans = min(tot,ans);
	
	return sum + 1;
	
}

int main()
{
	memset(h,-1,sizeof(h));
	
	cin >> n;
	
	for(int i = 1;i <= n-1;++i)
	{
		int a,b;
		cin >> a >> b;
		add(a,b);
		add(b,a);
	}
	
	dfs(1);
	
	cout << ans << endl;
	
	return 0;
	
}

（5）图的遍历

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

using namespace std;

const int N = 1e5 + 10;

int ne[N],h[N],e[N],idx = 0;
int dist[N];
bool st[N];
int n,m;

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

void bfs(int u)
{
	st[u] = true;
	queue<int>q;
	q.push(1);
	
	while(q.size() != 0)
	{
		int t = q.front();
		q.pop();
		st[t] = true;
		
		for(int i = h[t]; i != -1; i = ne[i])
		{
			
			int j = e[i];
			
			if(st[j]) continue;
			
			if(dist[j] > dist[t] + 1) dist[j] = dist[t] + 1;
			q.push(j);
		}
		 
	}
	
	return;
	
}

int main()
{
	memset(h,-1,sizeof(h));
	memset(dist,0x3f,sizeof(dist));
	
	cin >> n >> m;
	
	for(int i = 1;i <= m; ++i){
		int a,b;
		cin >> a >> b;
		
		add(a,b);
	}
	
	dist[1] = 0;
	bfs(1);
	
	if(dist[n] == 0x3f3f3f3f) cout << "-1" << endl;
	else cout << dist[n] << endl;
	
    return 0;
}

（6）拓扑序列

#include<iostream>
#include<cstring>
#include<vector>
#include<queue>

using namespace std;

const int N = 1e5 + 10;

int in[N];
int h[N],ne[N],e[N],idx = 0;
bool st[N];

vector<int>ans;
int n,m;

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

void bfs()
{	
	
	queue<int>q;
	
	for(int i = 1;i <= n; ++i) if(in[i] == 0) q.push(i);
	
	while(q.size() != 0)
	{
		int t = q.front();
		q.pop();
		ans.push_back(t);
		
		for(int i = h[t];i != -1;i = ne[i])
		{
			int j = e[i];
			in[j]--;
			if(in[j] == 0) q.push(j);
		}
	} 
	
	return;
	
}


int main()
{
	memset(h,-1,sizeof(h));
	
	cin >> n >> m;
	
	
	for(int i = 1;i <= m;++i){
		
		int a,b;
		cin >> a >> b;
		
		in[b]++;
		
		add(a,b);
	}
	
	bfs(); 
	
	if(ans.size() != n) cout << "-1" << endl;
	else{
		for(int i = 0;i < n; ++i) cout << ans[i] << ' '; 
	}
	
	return 0;
	
}