codeforces Magic Five_magic star codeforces-优快云博客

本文介绍了一道计数魔数的问题，通过高效算法解决如何在字符串中删除部分字符以形成能被5整除的最大数量的数字。文章提供了一个具体的C++实现方案，并通过样例输入输出展示了算法的有效性。

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There is a long plate s containing n digits. Iahub wants to delete some digits (possibly none, but he is not allowed to delete all the digits) to form his "magic number" on the plate, a number that is divisible by 5. Note that, the resulting number may contain leading zeros.

Now Iahub wants to count the number of ways he can obtain magic number, modulo 1000000007 (10⁹ + 7). Two ways are different, if the set of deleted positions in s differs.

Look at the input part of the statement, s is given in a special form.

Input

In the first line you're given a string a (1 ≤ |a| ≤ 10⁵), containing digits only. In the second line you're given an integer k (1 ≤ k ≤ 10⁹). The plate s is formed by concatenating k copies of a together. That is n = |a|·k.

Output

Print a single integer — the required number of ways modulo 1000000007 (10⁹ + 7).

Sample test(s)

input
1256
1

output
4

input
13990
2

output
528

input
555
2

output
63

题是水题，然后我刚开始都是用递归写的，然后就是超内存了，然后自己的脑袋不转圈了，然后就是傻了，好长时间没写题了不知道该怎么弄了，然后看了别人的代码然后才反应过来有个用递归写的完全是没有必要的，而我只想到了递归，可能递归每次写着都比较简单，所以我习惯了吧，废话不多说，贴个代码好了！

#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<queue>
#include<stack>
#include<vector>
#include<climits>
#include<map>
using namespace std;
#define rep(i,n) for(int i=0; i<(n); i++)
#define repf(i,n,m) for(int i=(n); i<=(m); i++)//正循环的
#define repd(i,n,m) for(int i=(n); i>=(m); i--) //负循环的
#define fab(a) ((a)>0?(a):(0-(a)))
#define ll long long
#define arc(a) ((a)*(a))
#define inf 1000000007   //最大值的
#define exp 0.0000001     //浮点型的
#define N 200000    //记录开的数组
char s[N];
ll n,k,dp;
int len;
ll fun(ll m,int x)
{
    ll ans=1;
    while(x!=0)
    {
        if(x%2)
          ans=(ans*m)%inf;
        m=(m*m)%inf;
        x/=2;
    }
    return ans;
} 
ll rec(int x)
{ 
   if(x==0)
     return 1;
   if(x%2==0)
   {
      return (fun(dp,x/2)%inf+((1+fun(dp,x/2+1)%inf)*rec(x/2-1))%inf)%inf;
   }
   else
      return ((1+fun(dp,x/2+1)%inf)*rec(x/2))%inf;
}
int main()
{
    while(scanf("%s",s)!=EOF)
    {
        cin>>k;
        ll ans=0,m=1;
        len=strlen(s);
        repf(i,0,len-1)
        { 
          if(s[i]=='0' || s[i]=='5')
          {
              ans+=m;
              if(ans>inf) ans-=inf;
          } 
          m*=2; if(m>inf) m-=inf;
        }  
        dp=fun(2,len);
    //    cout<<dp<<endl;
    //    cout<<rec(k-1)<<endl;
        ans*=rec(k-1);
        if(k==0)
          ans=0;
        cout<<ans%inf<<endl;
    }
     return 0;
}