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
#include
#include
#include
#define LL long long
#define eps 1e-8
#define maxn 1010
#define mod 1000000007
#define inf 0x3f3f3f3f
#define mid(a,b) ((a+b)>>1)
#define IN freopen("in.txt","r",stdin);
using namespace std;
LL n,m;
LL quickmod(LL a,LL b,LL m)
{
LL ans = 1;
while(b){
if(b&1){
ans = (ansa)%m;
b--;
}
b/=2;
a = aa%m;
}
return ans;
}
int main(int argc, char const *argv[])
{
//IN;
while(scanf("%lld %lld", &n,&m) != EOF)
{
char tmp[80]; gets(tmp);
LL cnt = 0;
LL cur = 0;
for(int i=1; i<=n; i++) {
char c = getchar(); c -= '0';
cur = (cur*10LL + c) % m;
if(cur == 0LL) cnt++;
}
if(cur) {
printf("0\n");
continue;
}
LL ans = quickmod(2LL, cnt-1LL, mod);
printf("%lld\n", ans);
//cout << ans << endl;
}
return 0;
}