Codeforces Round #330 (Div. 2) B. Pasha and Phone 容斥定理

B. Pasha and Phone

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/595/problem/B

Description

Pasha has recently bought a new phone jPager and started adding his friends' phone numbers there. Each phone number consists of exactly n digits.

Also Pasha has a number k and two sequences of length n / k (n is divisible by k) a₁, a₂, ..., a_n / k and b₁, b₂, ..., b_n / k. Let's split the phone number into blocks of length k. The first block will be formed by digits from the phone number that are on positions 1, 2,..., k, the second block will be formed by digits from the phone number that are on positions k + 1, k + 2, ..., 2·k and so on. Pasha considers a phone number good, if the i-th block doesn't start from the digit b_i and is divisible by a_i if represented as an integer.

To represent the block of length k as an integer, let's write it out as a sequence c₁, c₂,...,c_k. Then the integer is calculated as the result of the expression c₁·10^k - 1 + c₂·10^k - 2 + ... + c_k.

Pasha asks you to calculate the number of good phone numbers of length n, for the given k, a_i and b_i. As this number can be too big, print it modulo 10⁹ + 7.

Input

The first line of the input contains two integers n and k (1 ≤ n ≤ 100 000, 1 ≤ k ≤ min(n, 9)) — the length of all phone numbers and the length of each block, respectively. It is guaranteed that n is divisible by k.

The second line of the input contains n / k space-separated positive integers — sequence a₁, a₂, ..., a_n / k (1 ≤ a_i < 10^k).

The third line of the input contains n / k space-separated positive integers — sequence b₁, b₂, ..., b_n / k (0 ≤ b_i ≤ 9).

Output

Print a single integer — the number of good phone numbers of length n modulo 10⁹ + 7.

Sample Input

6 2
38 56 49
7 3 4

Sample Output

HINT

题意

给你ai和bi，让你找到有多少个k位数，使得这个k位数不以bi开头且mod ai=0

处理n/k次，然后把所有的答案都乘起来

题解：

容斥定理，所有的方案数 - 以bi开头的就好了

代码

#include<iostream>
#include<stdio.h>
using namespace std;
#define maxn 100005
const int mod = 1e9 + 7;
long long a[maxn],b[maxn];
long long ten[20];
long long ans[maxn];
int main()
{
    int n,k;scanf("%d%d",&n,&k);
    for(int i=1;i<=n/k;i++)
        scanf("%lld",&a[i]);
    for(int i=1;i<=n/k;i++)
        scanf("%lld",&b[i]);
    ten[0]=1;
    for(int i=1;i<=10;i++)
        ten[i]=ten[i-1]*10;
    long long tmp1,tmp2,tmp3;
    for(int i=1;i<=n/k;i++)
    {
        tmp1 = (ten[k]-1)/a[i]+1;
        tmp2 = ((b[i]+1)*ten[k-1]-1)/a[i]+1;
        if(b[i]==0)ans[i]=tmp1-tmp2;
        else{
            tmp3 = ((b[i])*ten[k-1]-1)/a[i]+1;
            ans[i]=tmp1-tmp2+tmp3;
        }
    }
    long long Ans = 1;
    for(int i=1;i<=n/k;i++)
        Ans = (Ans * ans[i])%mod;
    printf("%lld
",Ans);
}