前言

练习了很多场cf都没有好好的总结了，以后还是是不是总结一下每一场出现的问题，无论是解题经验还是解题时的心态问题，都是十分重要的，要避免重复问题出现。

一、A. Div. 7

A题戳这里

题解：

#include <algorithm>
#include <cmath>
#include <cstring>
#include <iostream>

using namespace std;
typedef long long ll;
const int N = 2e6 + 10;

int main()
{
	int T;
	cin >> T;
	while (T--)
	{
		int n;
		cin >> n;
		if(n%7==0)
			cout << n << endl;
		else{
			for (int i = 0; i < 10;i++){
				n /= 10;
				n = n * 10 + i;
				if(n%7==0){
					cout << n << endl;
					break;
				}
			}
		}
	}
	return 0;
}

二、B. Minority

B题戳这里

题解：

#include <algorithm>
#include <cmath>
#include <cstring>
#include <iostream>

using namespace std;
typedef long long ll;
const int N = 2e6 + 10;

int main()
{
	int T;
	cin >> T;
	while (T--)
	{
		string s;
		cin >> s;
		int a = 0, b = 0;
		for (int i = 0; i < s.size();i++){
			if(s[i]=='0')
				a++;
			else
				b++;
		}
		if(a==b)
			cout << max(a - 1, 0) << endl;
		else
		cout << min(a, b) << endl;
	}
	return 0;
}

C. Kill the Monster

C题戳这里

题解：

#include <algorithm>
#include <cmath>
#include <cstring>
#include <iostream>

using namespace std;
typedef long long ll;
const int N = 2e6 + 10;

int main()
{
	int T;
	cin >> T;
	while (T--)
	{
		ll hc, dc, hm, dm;
		cin >> hc >> dc >> hm >> dm;
		ll k, w, a;
		cin >> k >> w >> a;
		int flag = 0;
		for (int i = 0; i <= k; i++)
		{
			int j = k - i;
			ll t1 = (hc + i * a) / dm, t2 = hm / (dc + j * w);
			if ((hc + i * a) % dm)
				t1++;
			if (hm % (dc + j * w))
				t2++;
			if (t1 >= t2)
			{
				flag = 1;
				break;
			}
		}
		if (flag)
			cout << "YES" << endl;
		else
			cout << "NO" << endl;
	}
	return 0;
}

D. Make Them Equal

D题戳这里

题解：

#include <algorithm>
#include <cmath>
#include <cstring>
#include <iostream>

using namespace std;
typedef long long ll;
const int N = 1e5 + 10;

ll w[N], dp[N];
int main()
{
	ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
	memset(w, 0x3f3f3f3f, sizeof(w));
	w[1] = 0;
	for (int i = 1; i <= 1000; i++)
	{
		for (int j = 1; j <= i; j++)
			w[i + i / j] = min(w[i + i / j], w[i] + 1);
	}
	int T;
	cin >> T;
	while (T--)
	{
		int n, k;
		cin >> n >> k;
		k = min(k, 12 * n);
		ll b[N], c[N];
		for (int i = 1; i <= n; i++)
		{
			cin >> b[i];
			b[i] = w[b[i]];
		}
		for (int i = 1; i <= n; i++)
			cin >> c[i];
		memset(dp, 0, sizeof(dp));
		for (int i = 1; i <= n; i++)
		{
			for (int j = k; j >= b[i]; j--)
			{
				dp[j] = max(dp[j], dp[j - b[i]] + c[i]);
			}
		}
		cout << dp[k] << endl;
	}
	return 0;
}