A. 在小小的日历里面数呀数呀数

xfxcy 的专属日期问题模板，公主请抄。

#include <bits/stdc++.h>
using namespace std;
#define ls u << 1
#define rs u << 1 | 1
#define LL long long
#define int long long
#define PII pair <int, int>
#define fi first
#define se second
#define pub push_back
#define pob pop_back
#define puf push_front
#define pof pop_front
#define lb lower_bound
#define ub upper_bound
#define i128 __int128
#define pcnt(x) __builtin_popcount(x)
#define mem(a,goal) memset(a, (goal), sizeof(a))
#define rep(x,start,end) for(int x = (start) - ((start) > (end)); x != (end) - ((start) > (end)); ((start) < (end) ? x ++ : x --))
#define aLL(x) (x).begin(), (x).end()
#define sz(x) (int)(x).size()
const int INF = 998244353;
const int mod = 1e9 + 7;
const int N = 100010;
int a[13] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
int b[13] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
bool check(int x)
{
    if(x % 4 == 0 && x % 100 != 0 || x % 400 == 0) return true;
    return false;
}
void solve()
{
    //2116
    //32872
    int res = 0;
    for(int i = 2025; i < 2116; ++ i)
    {
        if(check(i))//判断闰年用 b 数组多一天
        {
            for(int j = 1; j <= 12; ++ j)
            {
                for(int k = 1; k <= b[j]; ++ k)
                {
                    int num = i * 10000;
                    num += j * 100;
                    num += k;
                    if(num > 20250329) res ++;
                }

            }

        }
        else
        {
            for(int j = 1; j <= 12; ++ j)
            {
                for(int k = 1; k <= a[j]; ++ k)
                {
                    int num = i * 10000;
                    num += j * 100;
                    num += k;
                    if(num > 20250329) res ++;
                }

            }
        }
    }
    cout << res + 1 << '\n';//加上一月一日
}

signed main()
{
    int t = 1;
    while (t --) solve();
    return 0;
}

B. 好数

纯暴力

#include <iostream>
#include <string>
#include <vector>
#include <algorithm>
using namespace std;
int n, res;
int check(int x)
{
	int cnt = 1;//cnt % 2 == 0 or 1 判断当前位置是奇数位置还是偶数位置
	while(x)
	{
		if(cnt % 2 != x % 10 % 2) return 0;
		x /= 10;
		cnt ++;
	}
	return 1;
}
int main()
{
	cin >> n;
	for(int i = 1; i <= n; ++ i) if(check(i)) res ++;
	
	cout << res << '\n';
    return 0;
}

C. R 格式

第一章 高精度

#include <iostream>
#include <string>
#include <vector>
#include <algorithm>
using namespace std;
int n;
string s;
string cal(string & s, int n)
{
	string res = "";
	reverse(s.begin(), s.end());
	int t = 0;
	for(int i = 0; i < s.size(); ++ i)
	{
		if(s[i] != '.')
		{
			t += (s[i] - '0') * n;
			res += to_string(t % 10);
			t /= 10;
		}
		else res += '.';
	}
	while(t)
	{
		res += to_string(t % 10);
		t /= 10;
	}
	reverse(res.begin(), res.end());
	return res;
}
int main()
{
	cin >> n >> s;
	
	for(int i = 0; i < n; ++ i) s = cal(s, 2);
	int pos = s.find('.');
	vector<int> res;
	if(s[pos + 1] >= '5')//四舍五入
	{
		int num = 1;
		for(int i = pos - 1; i >= 0; -- i)
		{
			num = (s[i] - '0' + num);
			res.push_back(num % 10);
			num /= 10;	
		}
		if(num) res.push_back(num % 10);
		for(int j = res.size() - 1; j >= 0; -- j) cout << res[j];
	}
	else
	{
		for(int j = 0; j < s.size(); ++ j)
		{
			if(s[j] != '.')
			{
				cout << s[j];
			}
			else break;
		}
		return 0;
	}
    return 0;
}

D. 冶炼金属

设答案为 $v$, 因为能炼出 $b$ 个但炼不出 $b+1$ 个所以有 $b×v≤a$ 且 $a<(b+1)×v$ 即 $\frac{a}{b+1}<v⩽\frac a b$

最小值 $\frac{a}{b+1}+1$，最大值 $\frac a b$

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

using namespace std;

int main()
{
    int n;
    scanf("%d", &n);
    int minn = 1, maxx = 1e9;
    while (n -- )
    {
        int a, b;
        scanf("%d%d", &a, &b);
        minn = max(minn, a / (b + 1) + 1);
        maxx = min(maxx, a / b);
    }

    printf("%d %d\n", minn, maxx);
    return 0;
}