FOJ 1675 The Seventy-seven Problem

The Seventy-seven Problem

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Time Limit:2s	Memory limit:32M
Accepted Submit:20	Total Submit:36

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One day, daxia writes a big number N that can be divided by seventy-seven on the paper. But oaiei is careless; he spilt some ink on the number, so that the four continuous digits can not be read clearly. Now daxia asks you to calculate how large the number N could be. Input There are multiple test cases. For each test case, the first line contains a big integer N (10^5<=N<=10^1000000) which had lost four continuous digits, the lost four continuous digits are represented as "xxxx". Output For each test case, you should output the number N, so that N is the maximum number that can be divided by seventy-seven. Sample Input 1xxxx 1234567xxxx7654321 Sample output 19943 123456799337654321 Original: FOJ月赛-2008年12月 题目意思很简单就是叫你把xxxx替换成0000~9999中的一个数，使得数串%77=0,并且替换成的数是最大的~ 一开始可能会想到找规律，确实这题规律比较明显，可以用找规律的方法做出来，这里不累述了 下面要介绍的是通常的解法 首先必须知道一点数论的基本公式  (a-b) %c =0 -----> a%c=b%c 于是就可以先把xxxx变成9999。 1234567xxxx7654321--->123456799997654321 123456799997654321 % 77 =44 - ????0000000 如果能找到这样的数,该数%77= 123456799997654321 % 77 =44，那么只要把9999-????就可以了 很容易知道00660000000 % 77=44,并且????0000000在这种数中是最小的,所以呢就把9999 - 66 = 9933 so 答案出来了 于是第一步是先把xxxx->9999,求出这时候的结果mod ????后面的0的个数可以很容易的通过o(n)扫描得到,如果后面有K个0，那么枚举数i从0000~9999,计算 (i * 10^K) mod 77 一直枚举直到 (i * 10^K) mod 77 = mod,退出，改变9999->9999-i 于是就AC了 注意必须使用比较快速的算法计算a^b mod c