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3960：Binary Operation

题目描述
Consider a binary operation __poj_jax_start__\odot__poj_jax_end__\odot defined on digits 0 to 9. __poj_jax_start__\odot__poj_jax_end__\odot: {0, 1, ..., 9} × {0, 1, ..., 9} __poj_jax_start__\to__poj_jax_end__\to {0, 1, ..., 9}, such that 0 __poj_jax_start__\odot__poj_jax_end__\odot 0 = 0.
A binary operation __poj_jax_start__\otimes__poj_jax_end__\otimes is a generalization of __poj_jax_start__\odot__poj_jax_end__\odot to the set of non-negative integers, __poj_jax_start__\otimes__poj_jax_end__\otimes: __poj_jax_start__\mathbb{Z}__poj_jax_end__\mathbb{Z}0+ × __poj_jax_start__\mathbb{Z}__poj_jax_end__\mathbb{Z}0+ __poj_jax_start__\to__poj_jax_end__\to __poj_jax_start__\mathbb{Z}__poj_jax_end__\mathbb{Z}0+. The result of a __poj_jax_start__\otimes__poj_jax_end__\otimes b is defined in the following way: if one of the numbers a and b has fewer digits than the other in decimal notation, then append leading zeroes to it, so that the numbers are of the same length;
then apply the operation __poj_jax_start__\odot__poj_jax_end__\odot digit-wise to the corresponding digits of a and b.



Let us define __poj_jax_start__\otimes__poj_jax_end__\otimes to be left-associative, that is, a __poj_jax_start__\otimes__poj_jax_end__\otimes b __poj_jax_start__\otimes__poj_jax_end__\otimes c is to be interpreted as (a __poj_jax_start__\otimes__poj_jax_end__\otimes b) __poj_jax_start__\otimes__poj_jax_end__\otimes c.
Given a binary operation __poj_jax_start__\odot__poj_jax_end__\odot and two non-negative integers a and b, calculate the value of a __poj_jax_start__\otimes__poj_jax_end__\otimes (a + 1) __poj_jax_start__\otimes__poj_jax_end__\otimes (a + 2) __poj_jax_start__\otimes__poj_jax_end__\otimes ... __poj_jax_start__\otimes__poj_jax_end__\otimes (b - 1) __poj_jax_start__\otimes__poj_jax_end__\otimes b.
输入解释
The first ten lines of the input file contain the description of the binary operation __poj_jax_start__\odot__poj_jax_end__\odot. The i-th line of the input file contains a space-separated list of ten digits - the j-th digit in this list is equal to (i - 1) __poj_jax_start__\odot__poj_jax_end__\odot (j - 1).
The first digit in the first line is always 0.
The eleventh line of the input file contains two non-negative integers a and b (0 <= a <= b <= 1018).
输出解释
Output a single number – the value of a __poj_jax_start__\otimes__poj_jax_end__\otimes (a + 1) __poj_jax_start__\otimes__poj_jax_end__\otimes (a + 2) __poj_jax_start__\otimes__poj_jax_end__\otimes ... __poj_jax_start__\otimes__poj_jax_end__\otimes (b - 1) __poj_jax_start__\otimes__poj_jax_end__\otimes b without extra leading zeroes.
输入样例 
0 1 2 3 4 5 6 7 8 9
1 2 3 4 5 6 7 8 9 0
2 3 4 5 6 7 8 9 0 1
3 4 5 6 7 8 9 0 1 2
4 5 6 7 8 9 0 1 2 3
5 6 7 8 9 0 1 2 3 4
6 7 8 9 0 1 2 3 4 5
7 8 9 0 1 2 3 4 5 6
8 9 0 1 2 3 4 5 6 7
9 0 1 2 3 4 5 6 7 8
0 10
输出样例 
15
该题目是Virtual Judge题目，来自 北京大学POJ
题目来源 Northeastern Europe 2010
源链接： POJ-3960
最后修改于 2020-10-29T07:16:39+00:00 由爬虫自动更新
共提交 0 次
通过率 --%
时间上限 内存上限
3000 65536