SCU 3578 H1N1's Problem

　　　　　　　　　　　　　　　　　　　　　　　　　　　3578: H1N1's Problem

Description

H1N1 like to solve acm problems.But they are very busy, one day they meet a problem. Given three intergers a,b,c, the task is to compute a^(b^c))%317000011. 1412, ziyuan and qu317058542 don't have time to solve it, so the turn to you for help.

Input The first line contains an integer T which stands for the number of test cases. Each case consists of three integer a, b, c seperated by a space in a single line. 1 <= a,b,c <= 100000

Output For each case, print a^(b^c)%317000011 in a single line.

Sample Input 2 1 1 1 2 2 2

Sample Output 1 16

 

快速幂+欧拉函数

地址 http://cstest.scu.edu.cn/soj/problem.action?id=3578

因为317000011为质数，所以对他求欧拉函数就是本身减一。直接两次快速幂，一开始类型没注意用longlong结果TLE了好几次，也是教训吧。

 1 #include<iostream>

 2 #include<cstdio>

 3 #include<cstdlib>

 4 #include<cstring>

 5 #include<string>

 6 #include<queue>

 7 #include<algorithm>

 8 #include<map>

 9 #include<iomanip>

10 #include<climits>

11 #include<string.h>

12 #include<numeric>

13 #include<cmath>

14 #include<stdlib.h>

15 #include<vector>

16 #include<stack>

17 #include<set>

18 #define INF 1e7

19 #define MAXN 100010

20 #define maxn 1000010

21 #define Mod 1000007

22 #define N 1010

23 using namespace std;

24 typedef long long LL;

25 

26 LL quick_pow(LL a, LL n, LL mod)

27 {

28     LL sum = 1;

29     while (n)

30     {

31         if (n & 1) sum = (sum*a)%mod;

32         a = (a%mod*a%mod) % mod;

33         n >>= 1;

34     }

35     return sum%mod;

36 }    

37 

38 void run()

39 {

40     LL a, b, c, ans;

41     scanf("%lld%lld%lld",&a,&b,&c);

42     ans = quick_pow(a, quick_pow(b, c, 317000010), 317000011);

43     printf("%lld\n",ans);

44 }

45 

46 int main()

47 {

48     int T;

49     scanf("%d", &T);

50     while (T--)

51         run();

52     return 0;

53 }