zoj3950——How Many Nines（日期模拟）

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.

求输入的两个时间段内有多少个9

先每年打表，接下来只要判断端点，时间会大大减少

#include <iostream>
#include <cstring>
#include <string>
#include <vector>
#include <queue>
#include <cstdio>
#include <set>
#include <math.h>
#include <algorithm>
#include <queue>
#include <iomanip>
#include <map>
#define INF 0x3f3f3f3f
#define MAXN 1005
#define Mod 99999999
using namespace std;
int num[10005];
int count9(int n)
{
int cnt=0;
while(n!=0)
{
if(n%10==9)
cnt++;
n/=10;
}
return cnt;
}
bool isleap(int y)
{
if((y%4==0&&y%100!=0)||y%400==0)
return true;
return false;
}
void Init()
{
int sum=0;
for(int i=2000; i<=9999; ++i)
{
if(isleap(i))
sum=count9(i)*366;
else
sum=count9(i)*365;
sum+=30;
sum+=11*3;
if(isleap(i))
sum+=3;
else
sum+=2;
num[i]=sum;
}
}
int mon1[]= {0,31,28,31,30,31,30,31,31,30,31,30,31};
int mon2[]= {0,31,29,31,30,31,30,31,31,30,31,30,31};
int main()
{
Init();
int t;
scanf("%d",&t);
while(t--)
{
int y1,m1,d1,y2,m2,d2;
int ans=0;
scanf("%d%d%d%d%d%d",&y1,&m1,&d1,&y2,&m2,&d2);
if(y1==y2)
{
int days=0;
for(int i=m1+1; i<m2; ++i)
{
if(i==2&&isleap(y1))
ans+=3;
else if(i==2&&!isleap(y1))
ans+=2;
else if(i==9)
ans+=33;
else
ans+=3;
if(!isleap(y1))
days+=mon1[i];
else
days+=mon2[i];
}
if(m1==m2)
{
for(int i=d1; i<=d2; ++i)
{
ans+=count9(i);
days++;
if(m1==9)
ans++;
}
ans=ans+days*count9(y1);
printf("%d\n",ans);
}
else
{
if(!isleap(y1))
{
for(int i=d1; i<=mon1[m1]; ++i)
{
ans+=count9(i);
days++;
if(m1==9)
ans++;
}
}
else
{
for(int i=d1; i<=mon2[m1]; ++i)
{
ans+=count9(i);
days++;
if(m1==9)
ans++;
}
}
for(int i=1; i<=d2; ++i)
{
ans+=count9(i);
days++;
if(m2==9)
ans++;
}
ans=ans+days*count9(y1);
printf("%d\n",ans);
}
}
else
{
for(int i=y1+1; i<y2; ++i)
ans+=num[i];
int days1=0,days2=0;
for(int i=m1+1; i<=12; ++i)
{
if(i==2&&isleap(y1))
ans+=3;
else if(i==2&&!isleap(y1))
ans+=2;
else if(i==9)
ans+=33;
else
ans+=3;
if(!isleap(y1))
days1+=mon1[i];
else
days1+=mon2[i];
}
if(!isleap(y1))
{
for(int i=d1; i<=mon1[m1]; ++i)
{
ans+=count9(i);
days1++;
if(m1==9)
ans++;
}
}
else
{
for(int i=d1; i<=mon2[m1]; ++i)
{
ans+=count9(i);
days1++;
if(m1==9)
ans++;
}
}
ans=ans+days1*count9(y1);

for(int i=1; i<m2; ++i)
{
if(i==2&&isleap(y2))
ans+=3;
else if(i==2&&!isleap(y2))
ans+=2;
else if(i==9)
ans+=33;
else
ans+=3;
if(!isleap(y2))
days2+=mon1[i];
else
days2+=mon2[i];
}
for(int i=1; i<=d2; ++i)
{
ans+=count9(i);
days2++;
if(m2==9)
ans++;
}
ans=ans+days2*count9(y2);
printf("%d\n",ans);
}
}
return 0;
}