HDU1095 A hard puzzle

题目描述：

lcy gives a hard puzzle to feng5166,lwg,JGShining and Ignatius: gave a and b,how to know the a^b.everybody objects to this BT problem,so lcy makes the problem easier than begin.
this puzzle describes that: gave a and b,how to know the a^b's the last digit number.But everybody is too lazy to slove this problem,so they remit to you who is wise

题目分析：我们只需要求a^b的最后一位，显然由于a和b太大，只能找规律。

题目转化：1.最后一位即个位，那我们不妨将a mod 10，因为只有a的最后一位在起作用；

2.若我们将a再乘b下，肯定还是超时；必然还要继续找规律；显然剩下的a只有10种情况，那我们为什麽不一一列举了？事实证明：它们具有周期性，简单吧。（没事写写水题也不错。）

代码：

View Code

 1 #include"stdio.h"
 2 int main()
 3 {
 4     int a,b;
 5     while(scanf("%d%d",&a,&b)!=EOF)
 6     {
 7          int s=1;
 8          a%=10;
 9          if(a==0||a==1||a==5||a==6)  s=a;
10          else if(a==2)
11          {
12            b%=4;
13            if(b==0) s=6;
14            else if(b==1) s=2;
15            else if(b==2) s=4;
16            else  s=8;
17          }
18          else if(a==3)
19          {
20             b%=4;
21             if(b==0) ;
22             else if(b==1)  s=3;
23             else if(b==2)  s=9;
24             else if(b==3)  s=7;
25          }
26          else if(a==4)
27          {
28             b%=2;
29             if(b==0)  s=6;
30             else s=4;
31          }
32          else if(a==7)
33          {
34             b%=4;
35             if(b==0) s=1;
36             else if(b==1) s=7;
37             else if(b==2) s=9;
38             else s=3;
39          }
40          else if(a==8)
41          {
42             b%=4;
43             if(b==0) s=6;
44             else if(b==1)  s=8;
45             else if(b==2)  s=4;
46             else if(b==3)  s=2;
47          }
48          else
49          {
50             b%=2;
51             if(b==0) ;
52             else s=9;
53          }
54         printf("%d\n",s);
55     }
56 return 0;
57 }

2013-01-09 20:39:08