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.
思路是分年份相同与否两种情况，前者直接暴力计算，后者取时间段两头的时间和中间的时间分开计算。比赛的时候卡了两次。第一次是因为计算年份不同时中间年份里的9而tle，选择采取了预处理。第二次是tle之后的wa……是因为自己失误少算了月份里的9。还是一道比较复杂的水题，因为wa得很多也懒得把代码改美观了，又臭又长。
#include <iostream>
#include <string>
#include <cstdio>
#include <map>
#include <vector>
#include <set>
#include <algorithm>
using namespace std;
int day[2][12] = {{3,2,3,3,3,3,3,3,33,3,3,3}, {3,3,3,3,3,3,3,3,33,3,3,3}};
int sumday[2][12] = {{31,28,31,30,31,30,31,31,30,31,30,31}, {31,29,31,30,31,30,31,31,30,31,30,31}};
int aaa[10011], summ=0;
bool check(int x) {
if (x % 400 == 0 || (x % 100 != 0 && x % 4 == 0)) return true;
else return false;
}
int nine(int a) {
int num[4], ans = 0;
num[0] = a / 1000, num[1] = (a / 100) % 10, num[2] = (a / 10) % 10, num[3] = a % 10;
if (num[0] == 9) ans++;
if (num[1] == 9) ans++;
if (num[2] == 9) ans++;
if (num[3] == 9) ans++;
return ans;
}
void init(){
int i;
for (i = 2000; i <= 10011; i++) {
if (check(i)) {
summ += 66;
summ += nine(i) * 366;
}
else {
summ += 65;
summ += nine(i) * 365;
}
aaa[i]=summ;
}
return;
}
int main() {
int t;
init();
scanf("%d", &t);
while (t--) {
int y, yy, m, mm, d, dd;
scanf("%d%d%d%d%d%d", &y, &m, &d, &yy, &mm, &dd);
int sum = 0;
if (y == yy) {
if (m == mm) {
for (int i = d; i <= dd; i++) {
if (i == 9 || i == 19 || i == 29) sum++;
sum += nine(y) + nine(m);
}
}
else {
for (int i = d; i <= (check(y) ? sumday[1][m - 1] : sumday[0][m - 1]); i++) {
if (i == 9 || i == 19 || i == 29) sum++;
sum += nine(y) + nine(m);
}
for (int i = m + 1; i < mm; i++) {
if (check(y)) {
sum += day[1][i - 1];
sum += nine(y) * sumday[1][i - 1];
}
else {
sum += day[0][i - 1];
sum += nine(y) * sumday[0][i - 1];
}
}
for (int i = 1; i <= dd; i++) {
if (i == 9 || i == 19 || i == 29) sum++;
sum += nine(y) + nine(mm);
}
}
}
else {
for (int i = d; i <= (check(y) ? sumday[1][m - 1] : sumday[0][m - 1]); i++) {
if (i == 9 || i == 19 || i == 29) sum++;
sum += nine (y) + nine(m);
}
for (int i = m + 1; i <= 12; i++) {
if (check(y)) {
sum += day[1][i - 1];
sum += nine(y) * sumday[1][i - 1];
}
else {
sum += day[0][i - 1];
sum += nine(y) * sumday[0][i - 1];
}
}
sum += aaa[yy-1] - aaa[y];
for (int i = 1; i < mm; i++) {
if (check(yy)) {
sum += day[1][i - 1];
sum += nine(yy) * sumday[1][i - 1];
}
else {
sum += day[0][i - 1];
sum += nine(yy) * sumday[0][i - 1];
}
}
for (int i = 1; i <= dd; i++) {
if (i == 9 || i == 19 || i == 29) sum++;
sum += nine(yy) + nine(mm);
}
}
printf("%d\n", sum);
}
return 0;
}