[BOJ 11520] And Then There Was 5
View as PDFYour friend Mike is arguing with a classmate Chris over what the last digit in π is. (You and I know there is no such thing, but Chris simply isn't to be dissuaded.) Mike's argument is as flawed as it is entertaining: given any digit in π that you look at, either the number itself is a 5 or there is a 5 that occurs after that digit.</p>
Cute, but that same argument is true of any digit, at least for the portion of π that has been calculated. Luckily, you're not being asked to actually prove anything; Mike just wants you to use your "computer skills" to give him some sample data for him to use to support his case.
Your friend wants a large sampling of places of π and wants to show that his argument holds true. For each test case, you will be given an offset into the digits of π (P) and a single digit (D).
He wants to you find the Qth and Rth digits of π given that:
- Q is the lowest offset such that Q >= P and the Qth digit of π equals D
- R is the lowest offset such that R > Q and the Rth digit of π equals 5
For example, if P is 3 and D is 2, Q would be 6 and R would be 8, and the Qth and Rth digits of π would be 2 and 5 respectively.
P Q R Offset: 0 1 2 3 4 5 6 7 8 9 10 11 ... Value: 3 1 4 1 5 9 2 6 5 3 5 8 ...## 입력 형식
The first line contains the number of test cases N (1 <= N <= 10,000).
Each of the following N lines contains two integers:
- P (1 <= P <= 1,000,000), an offset into the digits of π
- D (0 <= D <= 9), a digit to find in π
For each test case, you are to output a single line containing the Qth and Rth digits of π as described above.
예제 입력
4
3 2
123456 6
999999 8
765432 7
예제 출력
2 5
6 5
8 5
7 5
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