Educational Codeforces Round 116 (Rated for Div. 2)

Educational Codeforces Round 116 (Rated for Div. 2)

A. AB Balance

题目大意：

给定一个字符串要求，最后的字符串ab的数量和ba的数量相同，然后输出最后的字符串，要求最小步骤

代码实现：

#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <vector>
using namespace std;
#define int long long
#define endl '\n'

void solve()
{
    string s;
    cin >> s;
    int n = s.size(), resab = 0, resba = 0;
    for (int i = 0; i < n - 1; i++) {
		if (s[i] == 'a') if (s[i + 1] == 'b') resab++;
		if (s[i] == 'b') if (s[i + 1] == 'a') resba++; 
    }
	if (resab == resba) {
		cout << s << endl;
		return ;
	} else {
		for (int i = 0; i < n - 1; i++) {
			cout << s[i];
		}
		if (s[n - 1] == 'a') cout << "b\n";
		else cout << "a\n";
	}
}
signed main()
{
#ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin), freopen("out.txt", "w", stdout);
#endif
    ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
    int T;
    cin >> T;
    // T = 1;
    while (T--) {
        solve();
    }
    return 0;
}

B. Update Files

题目大意：

其实就是首先是2的i - 1次幂，然后判断第一个k大于的地方在哪里，然后看这里的前缀和是否大于我们的n，然后判断哪一位，分类讨论即可

代码实现：

#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <cstring>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <stack>
using namespace std;
#define int long long
#define endl '\n'

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

const int N = 65;
int a[N], s[N], n, k;
void solve() {
	cin >> n >> k;
	n--;
	int pos = -1;
	for (int i = 1; i <= 65; i++) {
		if (a[i] >= k) {
			pos = i - 1;
			break;
		}
	}
	if (s[pos] > n) {
		for (int i = 1; i <= pos; i++) {
			if (s[i] == n) {
				cout << i << endl;
			} else if (s[i] > n) {
				cout << i << endl;
			}
		}
	} else if (s[pos] == n) {
		cout << pos << endl;
	} else if (s[pos] < n) {
		n -= s[pos];
		if (n % k == 0) {
			cout << pos + n / k << endl;
		} else {
			cout << pos + n / k + 1 << endl;
		}
	}
}
signed main()
{
#ifndef ONLINE_JUDGE
	freopen("in.txt", "r", stdin), freopen("out.txt", "w", stdout);
#endif
	ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
	int T;
	cin >> T;
	// T = 1;
	for (int i = 1; i <= 62; i++) {
		a[i] = qpow(2, i - 1);
		s[i] += s[i - 1] + a[i];
	}
	while(T--) {
		solve();
	}
	return 0;
}

C. Banknotes

题目大意：

要求用k个纸币组不成的最大整数，先把k + 1，然后判断两个纸币之间最多加多少，然后计算就可以了，记得开longlong

代码实现；

#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <cstring>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <stack>
using namespace std;
#define int long long
#define endl '\n'

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

int n, k;
void solve() {
	cin >> n >> k;
	k++;
	vector<int> a(n);
	for (auto &it : a) cin >> it;
	vector<int> b = a;
	for (auto &it : b) it = qpow(10, it);
	int res = 0;
	for (int i = 1; i < n; i++) {
		int tmp = (b[i] - b[i - 1]) / b[i - 1];
		if (tmp <= k) {
			k -= tmp;
			res += tmp * b[i - 1];
		} else {
			res += k * b[i - 1];
			k = 0;
			break;
		}
	}
	if (k) {
		res += k * b[n - 1];
	}
	cout << res << endl;
}
signed main()
{
	ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
	int T;
	cin >> T;
	// T = 1;
	while(T--) {
		solve();
	}
	return 0;
}