POJ 2478-Number theory training D problem Fermat theorem plus fast power

Description

Input

2

Output

2

        
 

This title means to give a number n, to divide it into 1 parts, 2 copies, 3 copies ... n copies (all of them are natural numbers) How many methods are there in total, and the data is very large and results in 1e9+7. Just beginning to see this problem a blank face, after the experience of learning predecessors, see some clues, after all, is related to mathematics, not to be seen. First, a list of n=1 to n=6 cases is found to satisfy an equation:
ans=2^ (n-1) mod (1e9+7), is coincidence or inevitable.
Originally this problem is high school probability contact, is a partition of the problem, said n something, put n-1 a partition is a question, get C (n-1,n-1), (c this symbol is a combination, the former in brackets is C of the subscript), put n-2 a partition, get C (n-1,n-2) ... Put a partition, that is, C (n-1,1), the combination of these combinations, is the two-item expansion, where the parentheses are the same (because 1 of any time 1). So the preceding equation is not accidental. Enclosed is a two-item theorem:
(a+b) ^n= (n from 0 to N) C (n,k) a^ (n-k) *b^k. Know what to count and how to go after. Input data is large, can only use string input, because it is the operation of Exponentiation, think of the fast power, just Chen teacher to the template, take to change into Longlong can be used. Considering that the power data is very large, we should use the Fermat theorem to power down.



Above is the Fermat theorem, the string to be recycled to take the remainder, with the rule of taking the remainder:
(A + b) mod n = ((a mod n) + (b mod n)) mod n;
(a–b) MoD n= ((a mod n)-(b mod n) + N) mod n;
AB MoD n= (a mod n) (b mod n) mod n;

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30 of each of the above.

#include <iostream>
#include <string.h>
#define MOD 1000000007
using namespace std;
Long Long int quckpow (long long int m,long long int n,long long int k)
{
    long long int b=1;
    while (n>0)
    {
        if (n&1) b= (b*m)%k;
        n=n>>1;
        M= (m*m)%k;
    } return b;
}

int main ()
{

    long long int sum,ans;
    Char c[100005];
    while (cin>>c)
    {       sum=0;
        int L=strlen (c);
        for (int i=0;i<l;i++)
            sum= (sum*10+ (c[i]-' 0 '))% (mod-1);
        Ans=quckpow (2,sum% (mod-1) -1,mod);
        cout<<ans<<endl;
    }
    return 0;
}