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2426:Remainder

题目描述
Coco is a clever boy, who is good at mathematics. However, he is puzzled by a difficult mathematics problem. The problem is: Given three integers N, K and M, N may adds ('+') M, subtract ('-') M, multiples ('*') M or modulus ('%') M (The definition of '%' is given below), and the result will be restored in N. Continue the process above, can you make a situation that "[(the initial value of N) + 1] % K" is equal to "(the current value of N) % K"? If you can, find the minimum steps and what you should do in each step. Please help poor Coco to solve this problem.

You should know that if a = b * q + r (q > 0 and 0 <= r < q), then we have a % q = r.
输入解释
There are multiple cases. Each case contains three integers N, K and M (-1000 <= N <= 1000, 1 < K <= 1000, 0 < M <= 1000) in a single line.

The input is terminated with three 0s. This test case is not to be processed.
输出解释
For each case, if there is no solution, just print 0. Otherwise, on the first line of the output print the minimum number of steps to make "[(the initial value of N) + 1] % K" is equal to "(the final value of N) % K". The second line print the operations to do in each step, which consist of '+', '-', '*' and '%'. If there are more than one solution, print the minimum one. (Here we define '+' < '-' < '*' < '%'. And if A = a1a2...ak and B = b1b2...bk are both solutions, we say A < B, if and only if there exists a P such that for i = 1, ..., P-1, ai = bi, and for i = P, ai < bi)
输入样例
2 2 2
-1 12 10
0 0 0
输出样例
0
2
*+

该题目是Virtual Judge题目,来自 北京大学POJ

题目来源 PKU Monthly,Wang Yijie

源链接: POJ-2426

最后修改于 2020-10-29T06:32:13+00:00 由爬虫自动更新

共提交 0

通过率 --%
时间上限 内存上限
3000 65536