Educational Codeforces Round 98 (Rated for Div. 2)

A

如果\(n = m\)，答案为\(2 \times n\)；如果\(n \ne m\)，答案为\(2 \times max(n,m) - 1\)



#include <bits/stdc++.h>

using namespace std;

int n, m;

int main()

{

	int __;

	scanf("%d", &__);

	while(__ -- )

	{

		scanf("%d%d", &n, &m);

		printf("%d\n", 2 * max(n, m) - (m != n));

	}

	return 0;

}

B

\(max(a_i) \times (n - 1) <= sum + ans\) ，并且\((n - 1) \mid (sum + ans)\)， \(ans\)如果是负数，转化成模\((n-1)\)意义下的正数即可.



#include <bits/stdc++.h>

using namespace std;

typedef long long LL;

const int N = 1e5 + 20;

int n, a[N];

int main()

{

	int __;

	scanf("%d", &__);

	while(__ --)

	{

		scanf("%d", &n);

		LL sum = 0, maxn = 0;

		for(int i = 1; i <= n; ++ i)

		{

			scanf("%d", &a[i]);

			sum += a[i];

			maxn = max(maxn, (LL)a[i]);

		}

		LL res = maxn * (n - 1) - sum;

		if(res < 0) res = (res % (n - 1) + (n - 1)) % (n - 1);

		printf("%lld\n", res);

	}

	return 0;

}

C



#include <bits/stdc++.h>

using namespace std;

const int N = 2e5 + 20;

char str[N];

int main()

{

	int __;

	scanf("%d", &__);

	while(__ -- )

	{

		scanf("%s", str);

		int res = 0, a = 0, b = 0;

		for(int i = 0; str[i]; ++ i)

		{

			if(str[i] == '[') a ++;

			if(str[i] == ']' && a) a --, res ++;

			if(str[i] == '(') b ++;

			if(str[i] == ')' && b) b --, res ++;

		}

		printf("%d\n", res);

	}

	return 0;

}

D

预处理fib，求\(2^n\)的逆元即可



#include <bits/stdc++.h>

using namespace std;

typedef long long LL;

const int MOD = 998244353;

const int N = 2e5 + 10;

int n, f[N];

int pow_mod(int a, int b, int p)

{

	int res = 1;

	while(b)

	{

		if(b & 1) res = (LL)res * a % p;

		a = (LL)a * a % p;

		b >>= 1;

	}

	return res;

}

int main()

{

	f[1] = f[2] = 1;

	for(int i = 3; i < N; ++ i) f[i] = (f[i - 1] + f[i - 2]) % MOD;

	scanf("%d", &n);

	int res = (LL)f[n] * pow_mod(pow_mod(2, n, MOD), MOD - 2, MOD) % MOD;

	printf("%d\n", res);

	return 0;

}

2020.11.21

巴特西

Educational Codeforces Round 98 (Rated for Div. 2)

A

B

C

D

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