?

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.

Examples

Input 6 2
38 56 49
7 3 4 Output 8 Input 8 2
1 22 3 44
5 4 3 2 Output 32400

自己英語水平捉雞，這道題目看了半天硬是沒看懂

大意就是說有n為一串數字的長度，k為把一串數字分組后每一組中數字的個數
又有a【i】b【i】使得數字串滿足 ，每一組為a【i】的倍數，且不以b【i】開頭；

這是我開始寫的,對于b【i】不能開頭，我是遍歷判斷的.....（不要打我，我只會這個）
結果自然不行 #include<stdio.h> #include<math.h> int main(){long long undo=1;long long a[100000],b[100000],s=0;int i,j,n,k,d,mod;while(scanf("%d%d",&n,&k)!=EOF){d=pow(10,k)-1;mod=pow(10,k-1);for(i=0;i<n/k;i++)scanf("%I64d",&a[i]);for(i=0;i<n/k;i++)scanf("%I64d",&b[i]);for(i=0;i<n/k;i++){s+=d/a[i]+1;for(j=0;j*a[i]<=d;j++){ //最多9個數量級循環判斷，我怕不是個傻子 if(j*a[i]/mod==b[i])s--;}undo*=s;undo%=1000000007;s=0;}printf("%I64d\n",undo%1000000007);undo=1;}return 0; }

　看了大神的代碼，直接用算式算出不滿足要求2的數字個數，在用總數減去...

又想起新生賽中一道簽到題，等差數列求和，我用了循環，答案是直接套公式算，速度肯定快

所以做題時可以直接用算式算出來的，就盡量不要用循環獲其他的算，大道至簡

于是改之后...還是wa了 ，而且各種調試，比較結果都找不出錯

百度之，發現是pow（）的問題，這東西精度不夠，而且里面都是double型，做題就不要用

函數原型：double pow( double x, double y );

pow頭文件：math.h/cmath(C++中)功能：計算x的y次方返回值：x不能為負數且y為小數，或者x為0且y小于等于0，返回冪指數的結果。

?

ac的代碼：

1 #include<stdio.h> 2 #include<math.h> 3 long long power(int k){ 4 int s=1; 5 while(k--)s*=10; 6 return s; 7 } 8 int main(){ 9 long long undo=1; 10 long long a[100000],b[100000],s=0; 11 int i,j,n,k,d,mod; 12 while(scanf("%d%d",&n,&k)!=EOF){ 13 d=power(k); mod=d/10; 14 for(i=0;i<n/k;i++)scanf("%lld",&a[i]); 15 for(i=0;i<n/k;i++)scanf("%lld",&b[i]); 16 for(i=0;i<n/k;i++){ 17 s+=(d-1)/a[i]+1; //帶零，所有a[i]在該位數下的倍數個數 18 if(b[i]!=0) 19 s-=((b[i]+1)*mod-1)/a[i]-(b[i]*mod-1)/a[i]; //減去10*b[i]~10*（b[i]+1）-1 內ai倍數的個數 eg；30~39 20 else if(b[i]==0) 21 s-=(mod-1)/a[i]+1; //如果bi是0，減去所有0頭的ai倍數 22 undo*=s; 23 undo%=1000000007; 24 s=0; 25 } 26 printf("%lld\n",undo); 27 undo=1; 28 } 29 return 0; 30 }

總之：1.能用式子解決的問題，就不要用循環獲判斷來解決

? ? ? ? ? ?2.能不用pow（），就不用pow（）。

轉載于:https://www.cnblogs.com/-ifrush/p/10104272.html

總結

以上是生活随笔為你收集整理的W - Pasha and Phone CodeForces - 595B （收益颇丰的数学题的全部內容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網站內容還不錯，歡迎將生活随笔推薦給好友。