比赛链接

A

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
#include <bits/stdc++.h>
using namespace std;

#define IO ios::sync_with_stdio(0);cin.tie(0);
#define endl '\n'
typedef long long ls;
const int N = 4e5+10;




void sol(){
	int x; cin >> x;
	int p = sqrt(x);
	if (p * (p + 1) == x)
		cout << "YES" <<endl;
	else
		cout << "NO" << endl;
}


int main(){
	IO
	sol();
}

B

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
#include <bits/stdc++.h>
using namespace std;

#define IO ios::sync_with_stdio(0);cin.tie(0);
#define endl '\n'
typedef long long ls;
const int N = 2e5 + 10;

int t;
int n, a[N];

void sol(){
	cin >> n;
	for (int i = 1; i <= n; i++) cin >> a[i];
	for (int i = 1; i < n; i++)
		for (int j = i +1; j <= n; j++)
			if (__gcd(a[i], a[j]) > 1){
				cout << a[i] << ' ' << a[j] << endl;
				return;
			}
	cout << -1 <<endl;
}


int main(){
	IO
	cin >> t;
	while (t--){
		sol();
	}
}

C

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
#include <bits/stdc++.h>
using namespace std;

#define IO ios::sync_with_stdio(0); cin.tie(0);
#define endl '\n'

// patternStart=0 表示模板 P = 0101...
// patternStart=1 表示模板 P = 1010...
// 我们抽出所有 s[i] == P[i] 的位置，形成 mid，求 mid 的最大绝对子段和
int calc(const string &s, int patternStart) {
    int mx = 0, mn = 0;      // 最大/最小子段和
    int curMax = 0, curMin = 0;

    int n = (int)s.size();
    for (int i = 0; i < n; i++) {
        char expected = ((i & 1) == 0 ? (patternStart ? '1' : '0')
                                      : (patternStart ? '0' : '1'));
        if (s[i] == expected) {
            int val = (s[i] == '1') ? 1 : -1; // '1'->+1, '0'->-1

            // Kadane 最大子段和
            curMax = max(0, curMax + val);
            mx = max(mx, curMax);

            // Kadane 最小子段和
            curMin = min(0, curMin + val);
            mn = min(mn, curMin);
        }
    }
    return max(mx, -mn);
}

int main() {
    IO
    int T;
    cin >> T;
    while (T--) {
        int n;
        string s;
        cin >> n >> s;

        // 变成 1010...：抽 P=0101... 的匹配位
        // 变成 0101...：抽 P=1010... 的匹配位
        int ans = min(calc(s, 0), calc(s, 1));
        cout << ans << endl;
    }
    return 0;
}

F

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
#include <bits/stdc++.h>
using namespace std;

#define IO ios::sync_with_stdio(0);cin.tie(0);
#define endl '\n'
typedef long long ls;
const int N = 510;

//int atx[13] = {-2,-1,-1,-1,0,0,0,0,0,1,1,1,2};
//int aty[13] = {0,-1,0,1,-2,-1,0,1,2,-1,0,1,0};

void sol(){
	int n; cin >> n;
	cout << n - 1 + n / 5 << endl;
}


int main(){
	IO
    int t;
	cin >> t;
	while (t--){
	sol();
	}
}

G

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
#include <bits/stdc++.h>
using namespace std;

#define IO ios::sync_with_stdio(0);cin.tie(0);
#define endl '\n'
typedef long long ls;
const int N = 1e5 + 10;

int t;


void sol(){
	int n, m; cin >> n >> m;
	int a[N], b[N];
	ls sum1 = 0; ls sum2 = 0;
	for (int i = 1 ; i<= n; i++){
		cin >> a[i];
		sum1 += a[i];
	}
	for (int i = 1; i <= m; i++){
		cin >> b[i];
		sum2 += b[i];
	}
	if (sum1 == sum2){
		cout << 1 << endl;
		return;
	}
	else if (sum1 < sum2){
		sort(b + 1, b + m + 1);
		int k = 0;
		for (int i = m; i >= 1; i--)
		{
			sum2 -= b[i];
			k++;
			if (sum2 <= sum1)
				break;
		}
		cout << k << endl;
	}
	else{
		sort(a + 1, a + n + 1);
		int k = 0;
		for (int i = n; i >= 1; i--)
		{
			sum1 -= a[i];
			k++;
			if (sum2 >= sum1)
				break;
		}
		cout << k << endl;
	}
}


int main(){
	IO
	cin >> t;
	while (t--)
		sol();
}

H

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
#include <bits/stdc++.h>
using namespace std;

#define IO ios::sync_with_stdio(0);cin.tie(0);
#define endl '\n'
typedef long long ls;
const int N = 2e5 + 10;

void sol(){
	double xa, ya, xb, yb; cin >> xa >> ya >> xb >> yb;
	if (xa == xb){
		printf("%.8lf", 4.0 / fabs(ya-yb)+fabs(xa));
		return;
	}
	if (ya == yb){
		if (fabs(xa-xb)*fabs(ya)==4)
			printf("0"); 
		else
			printf("no answer");
		return;
	}
	double lenab = (double)sqrt((double)(xa-xb)*(xa-xb)+(double)(ya-yb)*(ya-yb));
	double h = 4.0 / lenab;
	double k = (double)(yb-ya)/(xb-xa);
	double b = ya - k * xa;
	printf("%.8lf", (h*sqrt(1+k*k)-b)/k);
}


int main(){
	IO
//	cin >> t;
//	while (t--){
	sol();
//	}
}

J

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
#include <bits/stdc++.h>
using namespace std;

#define IO ios::sync_with_stdio(0);cin.tie(0);
#define endl '\n'
typedef long long ls;
const int N = 1e5 + 10;

int t;

 ls qs(ls a, ls p){
	ls ans = 1;
	while (p){
		if (p % 2 == 1){
			ans *= a;
		}
		p >>= 1;
		a *= a;
	}
	return ans;
 }


void sol(){
	ls n; cin >> n;
	int q; cin >> q;
	while (q--){
		ls x; cin >> x;
		ls l = 1, r = 68;
		while (l < r){
			ls mid = l + (r - l) / 2;
			if (qs(2, mid) > x)
				r = mid;
			else
				l = mid + 1;
		}
		if (qs(2, l) - 1 <= n)
			cout << qs(2, l - 1) << endl;
		else{
			cout << n - qs(2, l - 1) + 1 << endl;
		}
	}
}


int main(){
	IO
	cin >> t;
	while (t--)
		sol();
}