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.

【Link】

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=5569

【题解】

啦啦啦博客督促我补题耶耶耶

nine[ ]记录2000-01-01到9999-12-31的每天累计的9的数量

【代码】

#include<bits/stdc++.h>
using namespace std;
int nine[10000][14][35];  
int month1[13]={0,31,29,31,30,31,30,31,31,30,31,30,31};  //闰年
int month2[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
int f(int y)
{
    if(y%400==0||(y%4==0&&y%100!=0))
        return 1;
    return 0;
}
int cul(int n)
{
    int ans=0;
    while(n)
    {
        if((n%10)==9)
            ans++;
        n/=10;
    }
    return ans;
}
int fun()
{
    int y,yy,m,mm,d;
    for(y=2000;y<=9999;y++)
    {
        nine[y][0][0]+=nine[y-1][12][31];
        yy=cul(y);  
        if(f(y))
            for(m=1;m<=12;m++)
            {
                nine[y][m][0]+=nine[y][m-1][month1[m-1]];
                mm=cul(m);  
                for(d=1;d<=month1[m];d++)
                {
                    nine[y][m][d]=nine[y][m][d-1]+yy+mm;
                    if(d==9||d==19||d==29)
                        nine[y][m][d]++;
                }
            }
        else
            for(m=1;m<=12;m++)
            {
                nine[y][m][0]+=nine[y][m-1][month2[m-1]];
                mm=cul(m);
                for(d=1;d<=month2[m];d++)
                {
                    nine[y][m][d]=nine[y][m][d-1]+yy+mm;
                    if(d==9||d==19||d==29)
                        nine[y][m][d]++;

                }
            }
    }
}
int  main()
{
    fun();
    int n,y1,m1,d1,y2,m2,d2;
    scanf("%d",&n);
    while(n--)
    {
        scanf("%d%d%d%d%d%d",&y1,&m1,&d1,&y2,&m2,&d2);
        printf("%d\n",nine[y2][m2][d2]-nine[y1][m1][d1-1]);
    }
    return 0;
}