Niuke Xiaobai month race 52

Niuke Xiaobai moon race 52

Problem A

Discuss and judge directly ,7 Point every time to , Eight o'clock 1 branch ~5 Minute late , Other absenteeism .

Problem B

simulation , Press 10 The rule of digits is to try whether it can be carried . Notice when you see 5000 Can be carried into 10000 .

Problem C

Prefix and maintenance interval are accumulated , Finally, see which part has the least accumulation times , At this time, when all the other instructions except this instruction are wrong, the number of wrong instructions is the most .

Problem D

Tree array & ST surface

A round table , So we should increase to 2 times . The value indicated by the question can only take one arc at a time （ It can also be a whole circle ）, Start with each cake , Use two points to find out which cake you can eat when you are full , And then use ST Table find the maximum value of cream in this range . Time complexity $O(n\log n)$

This method has gone too far , But I was a little puzzled when I was writing the solution to the question , It seems that such enumeration can not guarantee that the cream group with the highest and lowest amount of cream does not include other higher cream ？ It seems that I can't express myself clearly .

Problem E

At first, I thought about balancing numbers , Copy a template and WA 了 , Because the contribution of the elements of the group added later is not considered .

The positive solution of the problem is ： Two points of addition and exclusion

Find out the contribution of demand value in all numbers , Then subtract the contribution of demand value in this group .

Problem F

Pack in groups , Didn't do it .

Code：

Problem A

// Good Good Study, Day Day AC.
#include <iostream>
#include <stdio.h>
#include <cstdio>
#include <stdlib.h>
#include <string>
#include <string.h>
#include <cstring>
#include <math.h>
#include <cmath>
#include <queue>
#include <deque>
#include <stack>
#include <vector>
#include <map>
#include <algorithm> 
#include <unordered_map>
#include <unordered_set>
#define ffor(i,a,b) for(int i=(a) ;i<=(b) ;i++)
#define rrep(i,a,b) for(int i=(a) ;i>=(b) ;i--)
#define mst(v,s) memset(v,s,sizeof(v))
#define IOS ios::sync_with_stdio(false),cin.tie(0)
#define ll long long
#define INF 0x7f7f7f7f7f7f7f7f
#define inf 0x7f7f7f7f
#define PII pair<int,int>
#define int long long

using namespace std;



int n, T = 1;

void ready()
{
    
	int a = 0, b = 0;
	cin >> n;
	while (n--) {
    
		string st;
		cin >> st;
		if (st[0] == '7') continue;
		if (st[0] == '9') b++;
		if (st[0] == '8') {
    
			int t = (st[2] - '0') * 10 + st[3] - '0';
			if (t <= 5 && t != 0) a++;
			else if (t > 5) b++;
		}
	}
	cout << a << ' ' << b << '\n';
}


void work()
{
    

}

signed main()
{
    
	IOS;
	cin>>T;
	while (T--) {
    
		ready();
		work();
	}
	return 0;
}

Problem B

// Good Good Study, Day Day AC.
#include <iostream>
#include <stdio.h>
#include <cstdio>
#include <stdlib.h>
#include <string>
#include <string.h>
#include <cstring>
#include <math.h>
#include <cmath>
#include <queue>
#include <deque>
#include <stack>
#include <vector>
#include <map>
#include <algorithm> 
#include <unordered_map>
#include <unordered_set>
#define ffor(i,a,b) for(int i=(a) ;i<=(b) ;i++)
#define rrep(i,a,b) for(int i=(a) ;i>=(b) ;i--)
#define mst(v,s) memset(v,s,sizeof(v))
#define IOS ios::sync_with_stdio(false),cin.tie(0)
#define ll long long
#define INF 0x7f7f7f7f7f7f7f7f
#define inf 0x7f7f7f7f
#define PII pair<int,int>
#define int long long

using namespace std;



int n, T = 1;

void ready()
{
    
	cin >> n;
}


void work()
{
    
	int t = 1, ans = n;
	if (ans < 10) {
    
		if (ans >= 5) ans = 10;
	}
	while (n >= t) {
    
		t *= 10;
		int b = n % t; 
		if (b >= 10) while (b >= 10) b = b / 10;
		n = n / t * t;
		if (b >= 5) n += t;
		//cout << "b=" << b << '\n';
		ans = max(ans, n);
		//cout << "n= " <<n<< '\n';
		//cout << "t=" << t << '\n'<<'\n';
	}
	cout << ans << '\n';
}

signed main()
{
    
	IOS;
	cin>>T;
	while (T--) {
    
		ready();
		work();
	}
	return 0;
}

Problem C

// Good Good Study, Day Day AC.
#include <iostream>
#include <stdio.h>
#include <cstdio>
#include <stdlib.h>
#include <string>
#include <string.h>
#include <cstring>
#include <math.h>
#include <cmath>
#include <queue>
#include <deque>
#include <stack>
#include <vector>
#include <map>
#include <algorithm> 
#include <unordered_map>
#include <unordered_set>
#define ffor(i,a,b) for(int i=(a) ;i<=(b) ;i++)
#define rrep(i,a,b) for(int i=(a) ;i>=(b) ;i--)
#define mst(v,s) memset(v,s,sizeof(v))
#define IOS ios::sync_with_stdio(false),cin.tie(0)
#define ll long long
#define INF 0x7f7f7f7f7f7f7f7f
#define inf 0x7f7f7f7f
#define PII pair<int,int>
#define int long long

using namespace std;


const int N = 1e6 + 6;
int n, T = 1, m;
int sum[N];

void ready()
{
    
	cin >> n >> m;
	ffor(i,1,m) {
    
		int op, x, y;
		cin >> op;
		if (op == 1) {
    
			cin >> x >> y;
			sum[x]++; sum[y + 1]--;
		}
		if (op == 2) {
    
			cin >> x;
			sum[x]++;
		}
		if (op == 3) {
    
			cin >> x;
			sum[1]++; sum[x+1]--;
		}
	}
	int ans = inf, cnt = 0;
	ffor(i, 1, n) {
    
		sum[i] += sum[i - 1];
		ans = min(sum[i], ans);
	}
// ffor(i, 1, n) cout << sum[i] << ' '; cout << '\n';
	cout << m - ans << ' ';
	ffor(i, 1, n) {
    
		if (sum[i] == ans) {
    
			cnt++;
		// cout << "i=" << i << '\n';
		}
	}
	cout << cnt;
}


void work()
{
    

}

signed main()
{
    
	IOS;
	// cin>>T;
	while (T--) {
    
		ready();
		work();
	}
	return 0;
}

Problem D

// Good Good Study, Day Day AC.
#include <iostream>
#include <stdio.h>
#include <cstdio>
#include <stdlib.h>
#include <string>
#include <string.h>
#include <cstring>
#include <math.h>
#include <cmath>
#include <queue>
#include <deque>
#include <stack>
#include <vector>
#include <map>
#include <algorithm> 
#include <unordered_map>
#include <unordered_set>
#define ffor(i,a,b) for(int i=(a) ;i<=(b) ;i++)
#define rrep(i,a,b) for(int i=(a) ;i>=(b) ;i--)
#define mst(v,s) memset(v,s,sizeof(v))
#define IOS ios::sync_with_stdio(false),cin.tie(0)
#define ll long long
#define INF 0x7f7f7f7f7f7f7f7f
#define inf 0x7f7f7f7f
#define PII pair<int,int>
#define int long long

using namespace std;


const int N = 1e6 + 5;;
int n, T = 1, s;
int b[N], d[N][22];

int lowbite(int x)
{
    
	return x & (-x);
}

void add_in(int i, int num, int c[])
{
    
	while (i <= n * 2)
	{
    
		c[i] += num;
		i += lowbite(i);
	}
	return;
}

int get_sum(int x, int c[])
{
    
	int sum = 0;
	while (x)
	{
    
		sum += c[x];
		x -= lowbite(x);
	}
	return sum;
}

void ready()
{
    
	cin >> n >> s;
	ffor(i, 1, n) {
    
		int bi;
		cin >> bi;
		add_in(i, bi, b);
		add_in(i + n, bi, b);
	}
	ffor(i, 1, n) {
    
		int di;
		cin >> di;
		d[i][0] = d[n + i][0] = di;
	}
	ffor(j, 1, 19) {
    
		ffor(i, 1, 2 * n) {
    
			d[i][j] = max(d[i][j - 1], d[i + (1 << (j - 1))][j - 1]);
		}
	}
}

int get_max(int x, int y)
{
    
	int mid = log2(y - x + 1);
	return max(d[x][mid], d[y - (1 << mid) + 1][mid]);
}

void work()
{
    
	if (get_sum(n, b) < s) {
    
		cout << -1;
		return;
	}
	int ans = inf;
	for (int i = 1; i <= n; i++) {
    
		int l = i, r = n + i - 1, mid;
		while (l < r) {
    
			mid = l + r >> 1;
			if (get_sum(mid, b) - get_sum(i - 1, b) < s) l = mid + 1;
			else r = mid;
		}
		ans = min(ans, get_max(i, l));
	}
	cout << ans;
}

signed main()
{
    
	IOS;
	// cin>>T;
	while (T--) {
    
		ready();
		work();
	}
	return 0;
}

Problem E

// Good Good Study, Day Day AC.
#include <iostream>
#include <stdio.h>
#include <cstdio>
#include <stdlib.h>
#include <string>
#include <string.h>
#include <cstring>
#include <math.h>
#include <cmath>
#include <queue>
#include <deque>
#include <stack>
#include <vector>
#include <map>
#include <algorithm> 
#include <unordered_map>
#include <unordered_set>
#define ffor(i,a,b) for(int i=(a) ;i<=(b) ;i++)
#define rrep(i,a,b) for(int i=(a) ;i>=(b) ;i--)
#define mst(v,s) memset(v,s,sizeof(v))
#define IOS ios::sync_with_stdio(false),cin.tie(0)
#define ll long long
#define INF 0x7f7f7f7f7f7f7f7f
#define inf 0x7f7f7f7f
#define PII pair<int,int>
#define int long long

using namespace std;

const int N = 1e6 + 6;
const int mod = 998244353;

int n, T = 1, k;
vector<int> ve[N], alls;

void ready()
{
    
	cin >> n >> k;
	ffor(i, 1, n) {
    
		int s;
		cin >> s;
		ffor(j, 1, s) {
    
			int x;
			cin >> x;
			alls.push_back(x);
			ve[i].push_back(x);
		}
		sort(ve[i].begin(), ve[i].end());
	}
	sort(alls.begin(), alls.end());
}


void work()
{
    
	int ans = 0;
	ffor(i, 1, n) {
    
		for (auto item : ve[i]) {
    
			int x = k - item;
			ans += (alls.end() - lower_bound(alls.begin(), alls.end(), x)) - (ve[i].end() - lower_bound(ve[i].begin(), ve[i].end(), x));
		}
	}
	ans /= 2;
	cout << ans % mod;
}

signed main()
{
    
	IOS;
	// cin>>T;
	while (T--) {
    
		ready();
		work();
	}
	return 0;
}