hdu 4704 Sum

Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 757    Accepted Submission(s): 363

Problem Description

 

 

Sample Input

2

 

 

Sample Output

2

Hint

1. For N = 2, S(1) = S(2) = 1. 2. The input file consists of multiple test cases.

 

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std ;
typedef long long LL ;
int MM = 1000000006 ;
char s[100001] ;
LL pow_mod( LL a , int n )
{   int mm = MM+1 ;
    if( n == 0 ) return 1 ;
    LL ans = pow_mod( a , n/2 ) ;
    ans = ((ans))*ans%mm;
    if((n&1)) ans = a*ans%mm ;
    return ans%mm;
}
int mod()
{
    int i , n ;
    LL ans = 0 ;
    n = strlen(s) ;
    // 大数取模
    for( i = 0 ; i < n ;i++ )
    {
        ans = ( ans*10 + (s[i]-'0'))%MM;
    }
    return (int)ans-1 ;
}
int main()
{
    int ans ;
    while( scanf("%s" , s ) != EOF )
    { // 费马小定理 a^phip % p == 1 ;
      // 将次公式a^n % p == a^(n%phip+phip)%p ;
      ans = (int)pow_mod(2,mod()) ;
      printf("%d\n" , ans ) ;
    }
    return 0 ;
}