# Codeforces 327C Magic Five & Code

Source: Internet
Author: User
Tags pow

Actually, it's very watery. "Does not
The code is not long anyway

Ideas:
For a string, it is easy to know that for a certain I is 0 or 5, there is a 2^i "string starting from No. 0, the first bit OK"
A string of selection is also very easy to find, in order to find out each of the selected method, add up "set a string has a K-type selection method
For multiple strings, the second string of one I "for the first bit of the string" has 2^ (strlen (s) +i) selection, that is (2^i) x (2^strlen (s)) species selection method. In this case, the sum of the selected method can be extracted as (2^strlen (s)) K selection method
In this case, the sum of the selection of the M string can be sorted into the sum of the geometric progression, and the geometric progression formula is ai=k (2^ (strlen (s) x (i-1)))
According to the geometric progression summation formula we get M-string and tm= (1-2^ (strlen (s) *m))/(1-2^strlen (s))%mod

And then the math formula is done.
Division modulo = = Well, the first reaction is to find the inverse ... On the method of solving the inverse element I just forgot to "check it Out", Fermat's Little's theorem to find a inverse B:
Ab≡1 (mod p)
a^ (p-1) ≡1 (mod p) "P is prime"
Inverse b is a^ (p-2), very witty transformation

Fast power to seek TM

No more.

* * Less modulo a few times WA several, and then look at the data found that they have calculated a negative response to come up

``` #include <iostream> #include <stdio.h> #include <string.h> #define MOD
1000000007 #define LL-long long using namespace std;
int k,l,x,y;
LL CNT;
String A;
ll Pow (ll A,ll b) {ll r=1;
while (b) {if (b&1) R*=a,r%=mod;
A*=a,a%=mod;
b>>=1;
} return R;
int main (void) {cin>>a>>k;
L=a.size ();
for (int i=0;i<l;i++) if (a[i]== ' 0 ' | | | | a[i]== ' 5 ') Cnt+=pow (2,i), cnt%=mod;
X=pow (2,l);
Y=pow (X,K);
X= ((1-x)%mod+mod)%mod;
Y= ((1-y)%mod+mod)%mod;
cout<< (cnt* (Y*pow (x,mod-2))%mod)%mod<<endl;
return 0; }```
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