ZOj 3950 How Many Nines (日起模拟，打表，预处理)

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.

代码

#include<bits/stdc++.h>using namespace std;int a[2][13]={   0,31,28,31,30,31,30,31,31,30,31,30,31,//平年    0,31,29,31,30,31,30,31,31,30,31,30,31};int leap(int x)//判断该年是否为闰年{    if(x%400==0) return 1;    if(x%4==0&&x%100!=0) return 1;    return 0;}int c[10000],f[10000],k[10000];int x=65 ,y=66 ;//x 是平年一年65个九,包含19 ，29 09,9月，同理y是闰年void table(){    for(int i=9;i<=9999;i++)    {        int tmp=i;        while(tmp)        {            if(tmp%10==9)    c[i]++;     tmp=tmp/10;        }    }    for(int i=2000;i<=9999;i++) f[i] = leap(i) + f[i-1];// 前缀和求闰年的个数    for(int i=2000;i<=9999;i++) k[i] = k[i-1] + c[i]*(365+leap(i));//前缀和这一年有多少个九，比如2009 年365个九在2009里出现}int main(){    table(); int T; scanf("%d",&T);    while(T--)    {        int A1,B1,C1 ,A2,B2,C2,ans=0; scanf("%d%d%d%d%d%d",&A1,&B1,&C1,&A2,&B2,&C2);        if(A1==A2||A1+1==A2)        {            int Y=A1, M=B1,D=C1;            while(1)            {                ans+=c[Y]+c[M]+c[D];                if(Y==A2&&M==B2&&D==C2) break;                D++;                if(D>a[leap(Y)][M])                {                    M++; D=1;                    if(M>12)  M=1, Y++;                }            }        }        else        {            int Y=A1, M=B1, D=C1;            while(1)            {                ans+=c[Y]+c[M]+c[D];                if(Y==A1&&M==12&&D==31) break;                D++;                if(D>a[leap(Y)][M])                {                    M++,  D=1;                    if(M>12)   M=1, Y++;                }            }            Y=A2, M=1, D=1;            while(1)            {                ans+=c[Y]+c[M]+c[D];                if(Y==A2&&M==B2&&D==C2) break;                D++;                if(D>a[leap(Y)][M])                {                    M++,  D=1;                    if(M>12)   M=1,  Y++;                }            }            ans+=(k[A2-1]-k[A1])+x*((A2-1)-A1)+(f[A2-1]-f[A1]);        }        printf("%d\n",ans);    }    return 0;}