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.

Under two situations the player could score one point.

⋅1. If you touch a buoy before your opponent, you will get one point. For example if your opponent touch the buoy #2 before you after start, he will score one point. So when you touch the buoy #2, you won't get any point. Meanwhile, you cannot touch buoy #3 or any other buoys before touching the buoy #2.

⋅2. Ignoring the buoys and relying on dogfighting to get point. If you and your opponent meet in the same position, you can try to fight with your opponent to score one point. For the proposal of game balance, two players are not allowed to fight before buoy #2 is touched by anybody.

There are three types of players.

Speeder: As a player specializing in high speed movement, he/she tries to avoid dogfighting while attempting to gain points by touching buoys.
Fighter: As a player specializing in dogfighting, he/she always tries to fight with the opponent to score points. Since a fighter is slower than a speeder, it's difficult for him/her to score points by touching buoys when the opponent is a speeder.
All-Rounder: A balanced player between Fighter and Speeder.

There will be a training match between Asuka (All-Rounder) and Shion (Speeder).
Since the match is only a training match, the rules are simplified: the game will end after the buoy #1 is touched by anybody. Shion is a speed lover, and his strategy is very simple: touch buoy #2,#3,#4,#1 along the shortest path.

Asuka is good at dogfighting, so she will always score one point by dogfighting with Shion, and the opponent will be stunned for T seconds after dogfighting. Since Asuka is slower than Shion, she decides to fight with Shion for only one time during the match. It is also assumed that if Asuka and Shion touch the buoy in the same time, the point will be given to Asuka and Asuka could also fight with Shion at the buoy. We assume that in such scenario, the dogfighting must happen after the buoy is touched by Asuka or Shion.

The speed of Asuka is V1 m/s. The speed of Shion is V2 m/s. Is there any possibility for Asuka to win the match (to have higher score)?

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

8

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);
}