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
4 2 93 1763534
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.
一开始年份一年年搜TLE了，然后开个数组记录一下不用特判的年份就过了
1 #include <iostream>
2 #include<cstdio>
3 #include<cstring>
4 #include<algorithm>
5 #include<stack>
6 #include<map>
7 #include<set>
8 #include<queue>
9 #include<string>
10 #include<vector>
11 using namespace std;
12
13 int y1,y2,m1,m2,d1,d2;
14 long long res;
15 int day[2][13]={0,31,28,31,30,31,30,31,31,30,31,30,31,
16 0,31,29,31,30,31,30,31,31,30,31,30,31};
17
18 //判断年份有几个9
19 int f(int year)
20 {
21 int a;
22 int sum=0;
23 for(int i=1;i<=4;i++)
24 {
25 a=year%10;
26 if(a==9)
27 sum++;
28 year/=10;
29 }
30 return sum;
31 }
32
33 long long y[10000];
34
35 int main()
36 {
37 int T,i,flag_r,ynine;
38 long long temp;
39 y[1999]=0;
40 for(i=2000;i<=9999;i++)
41 {
42 //如果是闰年
43 if((i%400==0)||(i%4==0&&i%100!=0))
44 {
45 temp=f(i)*366/*年*/+30/*月*/+12*3;
46 }
47 //如果不是闰年
48 else
49 {
50 temp=f(i)*365/*年*/+30/*月*/+12*3-1;
51 }
52 y[i]=y[i-1]+temp;
53 }
54 while(~scanf("%d",&T))
55 {
56 while(T--)
57 {
58 scanf("%d%d%d%d%d%d",&y1,&m1,&d1,&y2,&m2,&d2);
59 res=0;
60 if(y1!=y2)//不同年份
61 {
62 //特判y1年
63 if((y1%400==0)||(y1%4==0&&y1%100!=0))
64 flag_r=1;
65 else
66 flag_r=0;
67 ynine=f(y1);
68 //特判m1月
69 for(i=d1;i<=day[flag_r][m1];i++)
70 {
71 res+=ynine;//年
72 if(m1==9)
73 res++;//月
74 if(i%10==9)
75 res++;//日
76 }
77 //从m1+1月到12月
78 for(i=m1+1;i<=12;i++)
79 {
80 if(i==2)
81 {
82 if(flag_r==1)
83 res+=3;
84 else
85 res+=2;
86 }
87 else
88 res+=3;
89 /*日*/
90 if(i!=9)
91 res+=ynine*day[flag_r][i]/*年*/+0/*月*/;
92 else
93 res+=ynine*day[flag_r][i]+30;
94 }
95 //
96 //判断从y1+1到y2-1年
97 res+=y[y2-1]-y[y1];
98 //特判y2
99 if((y2%400==0)||(y2%4==0&&y2%100!=0))
100 flag_r=1;
101 else
102 flag_r=0;
103 ynine=f(y2);
104 //从1月到m2-1月
105 for(i=1;i<=m2-1;i++)
106 {
107 if(i==2)
108 {
109 if(flag_r==1)
110 res+=3;
111 else
112 res+=2;
113 }
114 else
115 res+=3;
116 if(i!=9)
117 res+=ynine*day[flag_r][i]+0;
118 else
119 res+=ynine*day[flag_r][i]+30;
120 }
121 //特判m2月
122 for(i=1;i<=d2;i++)
123 {
124 res+=ynine;
125 if(m2==9)
126 res++;
127 if(i%10==9)
128 res++;
129 }
130 }
131 else if(m1!=m2)//同年不同月
132 {
133 if((y1%400==0)||(y1%4==0&&y1%100!=0))
134 flag_r=1;
135 else
136 flag_r=0;
137 ynine=f(y1);
138 //特判m1月
139 for(i=d1;i<=day[flag_r][m1];i++)
140 {
141 res+=ynine;
142 if(m1==9)
143 res++;
144 if(i%10==9)
145 res++;
146 }
147 //从m1+1月到m2-1月
148 for(i=m1+1;i<=m2-1;i++)
149 {
150 if(i==2)
151 {
152 if(flag_r==1)
153 res+=3;
154 else
155 res+=2;
156 }
157 else
158 res+=3;
159 if(i!=9)
160 res+=ynine*day[flag_r][i]+0;
161 else
162 res+=ynine*day[flag_r][i]+30;
163 }
164 //特判m2月
165 for(i=1;i<=d2;i++)
166 {
167 res+=ynine;
168 if(m2==9)
169 res++;
170 if(i%10==9)
171 res++;
172 }
173 }
174 else//同年同月
175 {
176 if((y1%400==0)||(y1%4==0&&y1%100!=0))
177 flag_r=1;
178 else
179 flag_r=0;
180 ynine=f(y1);
181 for(i=d1;i<=d2;i++)
182 {
183 if(m1==9)
184 res+=f(y1)+1;
185 else
186 res+=f(y1);
187 if(i%10==9)
188 res++;
189 }
190 }
191 printf("%lld\n",res);
192 }
193 }
194 return 0;
195 }