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
#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;

}