How Many Nines ZOJ - 3950 打表大法好

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-D₂are 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++.

<h4< dd="">Output

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

<h4< dd="">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

<h4< dd="">Sample Output

4
2
93
1763534

<h4< dd="">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.

这题一开始打算暴力模拟过去 结果写着写着不对劲了

后面扔给专门写模拟的队友写了

暴力打表过的

这样好写多了

 #include <cstdio>
 #include <cstring>
 #include <queue>
 #include <cmath>
 #include <algorithm>
 #include <set>
 #include <iostream>
 #include <map>
 #include <stack>
 #include <string>
 #include <vector>
 #define  pi acos(-1.0)
 #define  eps 1e-6
 #define  fi first
 #define  se second
 #define  lson l,m,rt<<1
 #define  rson m+1,r,rt<<1|1
 #define  bug         printf("******\n")
 #define  mem(a,b)    memset(a,b,sizeof(a))
 #define  fuck(x)     cout<<"["<<x<<"]"<<endl
 #define  f(a)        a*a
 #define  sf(n)       scanf("%d", &n)
 #define  sff(a,b)    scanf("%d %d", &a, &b)
 #define  sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
 #define  sffff(a,b,c,d) scanf("%d %d %d %d", &a, &b, &c, &d)
 #define  pf          printf
 #define  FRE(i,a,b)  for(i = a; i <= b; i++)
 #define  FREE(i,a,b) for(i = a; i >= b; i--)
 #define  FRL(i,a,b)  for(i = a; i < b; i++)
 #define  FRLL(i,a,b) for(i = a; i > b; i--)
 #define  FIN         freopen("DATA.txt","r",stdin)
 #define  gcd(a,b)    __gcd(a,b)
 #define  lowbit(x)   x&-x
 #pragma  comment (linker,"/STACK:102400000,102400000")
 using namespace std;
 typedef long long  LL;
 typedef unsigned long long ULL;
 const int INF = 0x7fffffff;
 const int mod = 1e9 + ;
 const int maxn = 1e6 + ;
 int a[] = {, , , , , , , , , , , };
 int check(int x) {
     if ((x % ) ==  || (x %  ==  && x %  != )) return ;
     return ;
 }
 int get9(int x) {
     int ans = ;
     while(x) {
         int cnt = x % ;
         if (cnt == ) ans++;
         x /= ;
     }
     return ans;
 }
 int ans[][][];
 int main() {
     int year = , yue = , day = , cnt = , ret;
     while(year < ) {
         yue = ;
         while(yue <= ) {
             day = ;
             if (yue ==  && check(year)) ret = a[yue - ] + ;
             else ret = a[yue - ];
             while(day <= ret ) {
                 int temp = get9(year) + get9(yue) + get9(day);
                 ans[year][yue][day] = cnt + temp;
                 cnt = ans[year][yue][day];
                 day++;
             }
             yue++;
         }
         year++;
     }
     int n, v1, m1, d1, v2, m2, d2;
     sf(n);
     for (int i =  ; i < n ; i++) {
         sfff(v1, m1, d1);
         sfff(v2, m2, d2);
         printf("%d\n", ans[v2][m2][d2] - ans[v1][m1][d1] + get9(v1) + get9(m1) + get9(d1));
     }
     return ;
 }