UVALive 7327 Digit Division （模拟）

Digit Division

题目链接：

http://acm.hust.edu.cn/vjudge/contest/127407#problem/D

Description

We are given a sequence of n decimal digits. The sequence needs to be partitioned into one or more contiguous subsequences such that each subsequence, when interpreted as a decimal number, is divisible by a given integer m. Find the number of different such partitions modulo 109+7. When determining if two partitions are different, we only consider the locations of subsequence boundaries rather than the digits themselves, e.g. partitions 2|22 and 22|2 are considered different.

Input

The input file contains several test cases, each of them as described below. The first line contains two integers n and m (1 ≤ n ≤ 300000, 1 ≤ m ≤ 1000000) — the length of the sequence and the divisor respectively. The second line contains a string consisting of exactly n digits.

Output

For each test case, output a single integer — the number of different partitions modulo 109 + 7 on a line by itself.

Sample Input

4 2 1246 4 7 2015

Sample Output

4 0
##题意： 求有多少种方式把一个数串划分成多段，使得每段组成的数字能被m整除.
##题解： 先从整体考虑，如果能被划分成满足条件的若干段，那么整个原串组成的数字必定满足条件. (若A,B都能被m整除，则AB=A*100...+B一定能被m整除) 再考虑把原串划分成两段，当且仅当某个前缀组成的数能被m整除时，原串才能被分成两段. 划成多段同理. 那么结果就是，求有多少个前缀能恰好被m整除. 若有m个(不包括末尾)，结果就是 2^m. (相当于枚举每个位置分割or不分割).
##代码： ``` cpp #include #include #include #include #include #include #include