乘法逆元+快速幂

唉...

#include <iostream>
#include <string.h>
#include <cstdio>
#include <queue>
#include <map>
#include <vector>
#include <string>
#include <cstring>
#include <algorithm>
#include <math.h>

#define SIGMA_SIZE 26
#pragma warning ( disable : 4996 )

using namespace std;
typedef long long LL;

inline LL LMax(LL a,LL b)    { return a>b?a:b; }
inline LL LMin(LL a,LL b)    { return a>b?b:a; }
inline int Max(int a,int b) { return a>b?a:b; }
inline int Min(int a,int b) { return a>b?b:a; }
inline int gcd( int a, int b ) { return b==0?a:gcd(b,a%b); }
inline int lcm( int a, int b ) { return a/gcd(a,b)*b; }  //a*b = gcd*lcm
const LL INF = 0x3f3f3f3f3f3f3f3f;
const LL mod  = 1000000007;
const int inf  = 0x3f3f3f3f;
const int maxk = 1e5+5;
const int maxn = 1e5+5;

int K;
char str[maxn];

//计算a的b次方取模
LL getm( LL a, LL b, LL m )
{
    LL ans = 1, base = a;
    while( b )  
    {  
        if ( b & 1 )
            ans = (ans * base) % m;
        base = ( base * base ) % m;
        b >>= 1;  
    }  
    return ans;
}

// 求a关于p的逆元
LL getn( LL a, LL p )
{ return getm( a, p-2, p ); }

int main()
{
    cin >> K;
    scanf( "%s", str );

    LL ans = 0;
    int len = strlen(str);

    LL inv, a1 = 0;
    LL p = 1, tmp = getm( 2, len, mod );
    LL sum = 0;



    p = getm( tmp, K, mod );
    inv = getn( tmp-1, mod );
    sum = ( ((p-1)%mod) * (inv%mod) ) %mod;

    for (int i = len - 1; i >= 0; i--)
    {
        if (str[i] == '0' || str[i] == '5')
        {
            a1 = getm( 2, i, mod );
            ans += (a1*sum)%mod;
            ans %= mod;
        }
    }

    printf( "%lld
", ans );

    return 0;
}