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

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

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

     return ;
 }