uva 11029(数学)

题意：a^b次方，让你取低三位和高三位。

思路：低三位用快速模幂，高三位可以取对数。

代码如下：

 1 /**************************************************
 2  * Author     : xiaohao Z
 3  * Blog     : http://www.cnblogs.com/shu-xiaohao/
 4  * Last modified : 2014-03-28 22:01
 5  * Filename     : uva_11029.cpp
 6  * Description     : 
 7  * ************************************************/
 8 
 9 #include <iostream>
10 #include <cstdio>
11 #include <cstring>
12 #include <cstdlib>
13 #include <cmath>
14 #include <algorithm>
15 #include <queue>
16 #include <stack>
17 #include <vector>
18 #include <set>
19 #include <map>
20 #define MP(a, b) make_pair(a, b)
21 #define PB(a) push_back(a)
22 
23 using namespace std;
24 typedef long long ll;
25 typedef pair<int, int> pii;
26 typedef pair<unsigned int,unsigned int> puu;
27 typedef pair<int, double> pid;
28 typedef pair<ll, int> pli;
29 typedef pair<int, ll> pil;
30 
31 const int INF = 0x3f3f3f3f;
32 const double eps = 1E-6;
33 ll ta, tb;
34 
35 ll a_b_MOD_c(ll a, ll b, ll c)
36 {
37     if(b==1) return a%c;
38     ll temp = a_b_MOD_c(a, b/2, c);
39     if(b%2 == 1) return (temp*temp*a)%c;
40     else return (temp*temp)%c;
41 }
42 
43 int main()
44 {
45 //    freopen("in.txt", "r", stdin);
46 
47     int T;
48     cin >> T;
49     while(T--){
50         cin >> ta >> tb;
51         double ansa = 100*pow(10, fmod(tb*log10(ta), 1));
52         cout << (int)ansa;
53         int ansb = a_b_MOD_c(ta, tb, 1000);
54         printf("...%03d\n", ansb);
55     }
56     return 0;
57 }
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posted @ 2014-03-28 23:11  张小豪  阅读(460)  评论(0)  编辑  收藏  举报