洛谷题目练习——二分+搜索+贪心+数学

一、二分

1.P1102

二分+排序

#include<bits/stdc++.h>
using namespace std;
const int N = 2e5+5;
long long  a[N]; 
int main(){
	long long n, c; cin>>n>>c;
	for(int i = 0; i < n ;i++){
		cin>>a[i];
	}
	sort(a, a + n);
	long long cnt = 0;
	for(int i =0 ; i <  n ; i++){
		long long target = a[i] + c;
		cnt += upper_bound(a , a+ n,target) - lower_bound(a,a+n,target);//第一个大于目标值的索引值减去第一个大于等于目标值的索引值，累加下来就是统计素有符合题意的个数。
	}
	cout<<cnt<<'\n';
	return 0;
}

2.P1182

二分+贪心

#include<bits/stdc++.h>
using namespace std;
using ll = long long;
const int N = 1e5 + 5;
ll a[N];
int n, m;
bool check(ll limit){
	ll sum= 0;
	ll cnt = 1;
	for(int i = 0; i < n; i++){
		if(sum + a[i] <= limit){
			sum += a[i];
		}
		else{
			cnt++;
			sum = a[i];
		}
	}
	return cnt <= m;
}
int main() {
	cin>>n>>m;
	ll l = 0, r = 0;
	for(int i =0 ; i<n; i++){
		cin>>a[i];
		l = max(l,a[i]);//左边界 
		r += a[i];//右边界 
	}
	while(l < r){
		ll mid = l +(r - l) / 2;
		if(check(mid)){
			r = mid;
		}
		else {
			l = mid + 1;
		}
	}
	cout<<l<<'\n';
	return 0;
}

3.P1873

二分答案

#include<bits/stdc++.h>
using namespace std;
const int N = 1e6 + 5;
using ll = long long;
ll a[N];
ll n,m;
bool check(ll mid){
	ll wood = 0;
	for(int i = 0; i< n; i++){
		if(mid < a[i]){
			wood += a[i] - mid;
		}
		if(wood >= m)
			return true;
	}
	return wood >= m;
}
 int main(){
 	cin>>n>>m;
 	ll l = 0, r = 0;
 	for(int i = 0; i < n; i++){
 		cin>>a[i];
 		r = max(r,a[i]);
	 }
	 ll ans = 0;
	 while(l <= r){
	 	ll mid = l + (r - l) /2;
	 	if(check(mid)){
	 		ans = mid;
	 		l = mid + 1;
		 }
		 else{
		 	r = mid - 1;
		 }
	 }
	 cout<<ans<<'\n';
 	return 0;
 }

二、搜索

DFS

1.P1451

#include<bits/stdc++.h>
using namespace std;
const int N = 105;
char grid[N][N];//定义的是字符串二维数组
bool vis[N][N];
int n, m;
void dfs(int x, int y){
	int dx[4] = {-1, 1, 0, 0 };
	int dy[4] = {0, 0, -1, 1 };
	vis[x][y] = true;
	for(int i = 0; i < 4; i++){
		int nx = x + dx[i];
		int ny = y + dy[i];
		if(nx >= 0 && nx < n &&ny >=0 &&ny < m){
			if(grid[nx][ny] != '0' && !vis[nx][ny]){
				dfs(nx,ny);
			}
		}
	}
	return;
}
int main(){
	cin>>n>>m;
	int cnt = 0;
	for(int i = 0; i < n; i++){
		string s;
		cin>>s;
		for(int j = 0; j <m; j++){
			grid[i][j] = s[j];
		}
	}
	for(int i = 0; i< n; i++){
		for(int j = 0; j < m ;j++){
			if(grid[i][j] != '0' && !vis[i][j])
			{
				dfs(i,j);
				cnt++;
			}
		}
	}
	cout<<cnt<<'\n';
	return 0;
}

BFS

1.P1443|马的遍历

适用于c++11版本：

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

int main() {
    int n, m, x, y;
    cin >> n >> m >> x >> y;
    x--; y--;

    vector<vector<int>> f(n, vector<int>(m, -1));
    queue<pair<int,int>> q;
    q.push({x,y});
    f[x][y] = 0;

    int dx[8] = {1,1,2,2,-1,-1,-2,-2};
    int dy[8] = {2,-2,1,-1,2,-2,1,-1};

    while(!q.empty()){
        pair<int,int> cur = q.front(); q.pop();
        int a = cur.first;
        int b = cur.second;
        for(int i = 0; i < 8; i++){
            int nx = a + dx[i];
            int ny = b + dy[i];
            if(nx >= 0 && nx < n && ny >= 0 && ny < m && f[nx][ny] == -1){
                f[nx][ny] = f[a][b] + 1;
                q.push({nx,ny});
            }
        }
    }

    for(auto &row : f){
        for(int i = 0; i < row.size(); i++){
            cout << row[i];
            if(i != row.size()-1) cout << " ";
        }
        cout << "\n";
    }

    return 0;
}

三、贪心

1.P1094

#include<bits/stdc++.h>
using namespace std;
const int N = 3e4+5;
int a[N];

int main(){
	int w,n ; cin>>w>>n;
	for(int i =0 ; i < n; i++) cin>>a[i];
	sort(a,a+n);
	int l = 0, r = n - 1;
	int cnt = 0;
	while(l <= r){
		if(a[l] + a[r] <= w){
			l++;
		}
		r--;
		cnt++;
	}
	cout<<cnt<<'\n';
	return 0;
}

2.P1803

#include<bits/stdc++.h>
using namespace std;
const int N = 1e6+5;
int a[N];
struct t{
	int start;
	int end;
};
bool cmp(const t &a,const t &b){
	return a.end < b.end;
}
int main(){
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	int n; cin>>n;
	
	vector<t> ts(n);
	for(int i =0; i <n ;i++){
		cin>>ts[i].start>>ts[i].end;
	}
	int cnt = 0;
	int l_end = -1;
	sort(ts.begin(),ts.end(),cmp);
	for(int i = 0;  i< n; i++){
		if(ts[i].start >= l_end){
			cnt++;
			l_end = ts[i].end;
		}
	}
	cout<<cnt<<'\n';
	return 0;
}

四、数学

埃氏筛

// 质数筛
for (int i = 2; i * i <= m; i++)
    if (isPrime[i])
        for (int j = i * i; j <= m; j += i)
            isPrime[j] = false;

// 前缀和
pre[i] = pre[i - 1] + isPrime[i];

// 查询
ans = pre[r] - pre[l - 1];

1.P1865

部分超时

#include<bits/stdc++.h>
using namespace std;
bool prime(int x){
	if(x < 2) return false;
	for(int i = 2; i*i <= x; i++){
		if(x % i ==0) return false;
	}
	return true;
}
int cnt;
int main(){
	int n,m; cin>>n>>m;
	int l, r;
	for(int i =0 ; i < n; i++){
		cin>>l>>r;
		if(l < 1 || r > m){
			cout<<"Crossing the line"<<'\n';
			continue;
		}
		 cnt = 0;
		for(int j = l; j <= r; j++){
			if(prime(j)){
				cnt++;
			}
		}
        cout<<cnt<<'\n';
	}
	
	return 0;
}

使用埃氏筛

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

int main() {
    int n, m;
    cin >> n >> m;
    
    // 1. 埃拉托斯特尼筛法预处理质数
    vector<bool> isPrime(m + 1, true);
    isPrime[0] = isPrime[1] = false;
    for (int i = 2; i * i <= m; i++) {
        if (isPrime[i]) {
            for (int j = i * i; j <= m; j += i) {
                isPrime[j] = false;
            }
        }
    }

    // 2. 前缀和，count[i]表示[1..i]质数数量
    vector<int> count(m + 1, 0);
    for (int i = 1; i <= m; i++) {
        count[i] = count[i - 1] + (isPrime[i] ? 1 : 0);
    }

    // 3. 查询每个区间
    int l, r;
    for (int i = 0; i < n; i++) {
        cin >> l >> r;
        if (l < 1 || r > m) {
            cout << "Crossing the line\n";
        } else {
            cout << count[r] - count[l - 1] << "\n";
        }
    }

    return 0;
}

2.P1029

#include<bits/stdc++.h>
using namespace std;
int gcd(int n, int m){
	if(m == 0) return n;
	return gcd(m, n % m);
}
int main(){
	int x, y;
	cin>>x>>y;
	int cnt = 0;
	for(int i = x; i <= y; i++){
		if(x * y % i == 0 && gcd(i,x * y / i) == x) cnt++;
	} 
	cout<<cnt<<'\n';
	return 0;
}