【ZOJ3950 The 17th Zhejiang University Programming Contest C】【简单日期模拟】How Many Nines 区间9个数计数

最新推荐文章于 2019-08-28 13:24:43 发布

原创最新推荐文章于 2019-08-28 13:24:43 发布 · 880 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#题库-ZOJ #数据结构-前缀和 #日期天数处理

题库-ZOJ 同时被 3 个专栏收录

31 篇文章

订阅专栏

数据结构-前缀和

11 篇文章

订阅专栏

日期天数处理

1 篇文章

订阅专栏

本篇介绍了一个算法问题，即计算两个日期之间所有日期中数字9出现的总次数，考虑了普通年份和闰年的情况，并提供了一种有效的前缀和计数方法。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

How Many Nines

Time Limit: 1 Second Memory Limit: 65536 KB

If we represent a date in the format YYYY-MM-DD (for example, 2017-04-09), do you know how many 9s will appear in all the dates between Y₁-M₁-D₁ and Y₂-M₂-D₂ (both inclusive)?

Note that you should take leap years into consideration. A leap year is a year which can be divided by 400 or can be divided by 4 but can't be divided by 100.

Input

The first line of the input is an integer T (1 ≤ T ≤ 10⁵), indicating the number of test cases. Then T test cases follow. For each test case:

The first and only line contains six integers Y₁, M₁, D₁, Y₂, M₂, D₂, their meanings are described above.

It's guaranteed that Y₁-M₁-D₁ is not larger than Y₂-M₂-D₂. Both Y₁-M₁-D₁ and Y₂-M₂-D₂ are between 2000-01-01 and 9999-12-31, and both dates are valid.

We kindly remind you that this problem contains large I/O file, so it's recommended to use a faster I/O method. For example, you can use scanf/printf instead of cin/cout in C++.

Output

For each test case, you should output one line containing one integer, indicating the answer of this test case.

Sample Input

4
2017 04 09 2017 05 09
2100 02 01 2100 03 01
9996 02 01 9996 03 01
2000 01 01 9999 12 31

Sample Output

Hint

For the first test case, four 9s appear in all the dates between 2017-04-09 and 2017-05-09. They are: 2017-04-09 (one 9), 2017-04-19 (one 9), 2017-04-29 (one 9), and 2017-05-09 (one 9).

For the second test case, as year 2100 is not a leap year, only two 9s appear in all the dates between 2100-02-01 and 2100-03-01. They are: 2017-02-09 (one 9) and 2017-02-19 (one 9).

For the third test case, at least three 9s appear in each date between 9996-02-01 and 9996-03-01. Also, there are three additional nines, namely 9996-02-09 (one 9), 9996-02-19 (one 9) and 9996-02-29 (one 9). So the answer is 3 × 30 + 3 = 93.

Author: WENG, Caizhi

Source: The 17th Zhejiang University Programming Contest Sponsored by TuSimple

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<string>
#include<ctype.h>
#include<math.h>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<bitset>
#include<algorithm>
#include<time.h>
using namespace std;
void fre() { freopen("c://test//input.in", "r", stdin); freopen("c://test//output.out", "w", stdout); }
#define MS(x, y) memset(x, y, sizeof(x))
#define ls o<<1
#define rs o<<1|1
typedef long long LL;
typedef unsigned long long UL;
typedef unsigned int UI;
template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b > a)a = b; }
template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b < a)a = b; }
const int N = 0, M = 0, Z = 1e9 + 7, inf = 0x3f3f3f3f;
template <class T1, class T2>inline void gadd(T1 &a, T2 b) { a = (a + b) % Z; }
int casenum, casei;
int mon[13] = { 0,31,0,31,30,31,30,31,31,30,31,30,31 }, s[2][13];
int pre[10000][13][32];
int sum[10000][13][32];
int leap(int y)
{
	if (y % 400 == 0)return 1;
	if (y % 100 == 0)return 0;
	if (y % 4 == 0)return 1;
	return 0;
}
inline int check(int y, int m, int d)
{
	int ret = 0;
	while (y)
	{
		ret += y % 10 == 9;
		y /= 10;
	}
	while (m)
	{
		ret += m % 10 == 9;
		m /= 10;
	}
	while (d)
	{
		ret += d % 10 == 9;
		d /= 10;
	}
	return ret;
}
void init(int y1, int m1, int d1, int y2, int m2, int d2)
{
	mon[2] = 28 + leap(y1);
	int tmp = 0;
	while (y1 != y2 || m1 != m2 || d1 != d2)
	{
		pre[y1][m1][d1] = tmp;
		tmp += check(y1, m1, d1);
		sum[y1][m1][d1] = tmp;
		if (++d1 > mon[m1])
		{
			d1 = 1;
			if (++m1 > 12)
			{
				m1 = 1;
				mon[2] = 28 + leap(++y1);
			}
		}
	}
}
int main()
{
	init(2000, 1, 1, 10000, 1, 1);
	scanf("%d", &casenum);
	for (casei = 1; casei <= casenum; ++casei)
	{
		int y1, m1, d1, y2, m2, d2;
		scanf("%d%d%d%d%d%d", &y1, &m1, &d1, &y2, &m2, &d2);
		int ans = sum[y2][m2][d2] - pre[y1][m1][d1];
		printf("%d\n", ans);
	}
	return 0;
}
/*
【题意】
给你两个日期，问你这两个日期之间有多少个数字'9'

【分析】
询问组数非常多，但是范围很狭小，于是直接做10000 * 365 的预处理
因为要做 r - (l - 1)的前缀和计数，所以直接使用两个数组，会更方便哦~
所以我就拿了这题的一血嘛~

*/