牛客练习赛

牛客练习赛30

A-回文日期：题目链接

一个很麻烦的模拟题，不过值得一做（很锻炼代码能力）。

#include<iostream>
#include<algorithm>
#include<string>
#include<cstring>
using namespace std;
typedef long long ll;
const int maxn = 2000 + 10;
int month[maxn] = { 0,1,2,3,4,5,6,7,8,9,10,11,12 };
int day[maxn] = { 0,31,28,31,30,31,30,31,31,30,31,30,31 };
bool runnian(int x)
{
	if (x % 4 == 0 && x % 100 != 0 || x % 400)
		return true;
	return false;
}
int main()
{
	ios::sync_with_stdio(false);
	string s, e;
	int T;
	cin >> T;
	while (T--)
	{
		cin >> s >> e;
		int sY = (s[0] - 48) * 1000 + (s[1] - 48) * 100 + (s[2] - 48) * 10 + (s[3] - 48);
		int eY = (e[0] - 48) * 1000 + (e[1] - 48) * 100 + (e[2] - 48) * 10 + (e[3] - 48);
		int sM = (s[5] - 48) * 10 + (s[6] - 48);
		int eM = (e[5] - 48) * 10 + (e[6] - 48);
		int sD = (s[8] - 48) * 10 + (s[8] - 48);
		int eD = (e[8] - 48) * 10 + (e[8] - 48);
		//cout << sY;
		int ans = 0;
		for (int i = sY; i<eY; i++) {
			if (runnian(i))day[2] = 29;
			else day[2] = 28;
			for (int j = sM; j <= 12; j++) {
				if (j == 12)sM = 1;
				for (int k = sD; k <= day[month[j]]; k++) {
					if (k == day[month[j]])sD = 1;
					int t1 = (k % 10) * 1000;
					int t2 = (k / 10) * 100;
					int t3 = (j % 10) * 10;
					int t4 = j / 10;
					if (i == t1 + t2 + t3 + t4)ans++;
				}
			}
		}
		for (int i = 1; i<month[eM]; i++) {
			for (int j = 1; j <= day[month[eM] - 1]; j++) {
				int t1 = (j % 10) * 1000;
				int t2 = (j / 10) * 100;
				int t3 = (i % 10) * 10;
				int t4 = i / 10;
				if (t1 + t2 + t3 + t4 == eY)ans++;
			}
		}
		int t3 = (month[eM] % 10) * 10;
		int t4 = month[eM] / 10;
		for (int i = 1; i <= eD; i++) {
			int t1 = (i % 10) * 1000;
			int t2 = (i / 10) * 100;
			if (t1 + t2 + t3 + t4 == eY)ans++;
		}
		cout << ans << endl;
	}
	return 0;
}

C-小K的疑惑：题目链接

暂时没看懂这个dfs。。。有机会再补