HDU4704(SummerTrainingDay04-A 欧拉降幂公式)

Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 3245    Accepted Submission(s): 1332

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.

 

Source

2013 Multi-University Training Contest 10

 

模型最终转换为求2^(b-1) mod (1e9+7)，根据费马小定理可得1e9+7的欧拉函数为1e9+6。根据欧拉降幂公式a^b = a^(b%phi(MOD)+phi(MOD)) mod MOD，用快速幂算出a^(b%phi(MOD)+phi(MOD))即为答案。

//2017-08-04
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define ll long long

using namespace std;

const int N = 100010;
const int MOD = 1000000007;
char str[N];

ll quick_pow(ll a, ll n){//快速幂
    ll ans = 1;
    while(n){
        if(n&1)ans = ans*a%MOD;
        a = a*a%MOD;
        n>>=1;
    }
    return ans;
}

int main()
{
    while(scanf("%s", str)!=EOF){
        ll num = 0;
        for(int i = 0; i < strlen(str); i++){//欧拉降幂
            num *= 10;
            num += str[i]-'0';
            num %= (MOD-1);
        }
        num -= 1;
        printf("%lld
", quick_pow(2, num));
    }

    return 0;
}