Problem Description
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.
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.
Input
There are mutiple test cases. Each test cases consists
of two numbers a and b(0<a,b<=2^30)
Output
For each test case, you should output the a^b's last
digit number.
Sample Input
7 66
8 800
Sample Output
9
6
1 #include <stdio.h> 2 3 int main(){ 4 int a; 5 int b; 6 int number; 7 char two[5]="2486"; 8 char three[5]="3971"; 9 char four[3]="46"; 10 char seven[5]="7931"; 11 char eight[5]="8426"; 12 char nine[3]="91"; 13 14 while(scanf("%d%d",&a,&b)!=EOF){ 15 number=a%10; 16 17 if(number==0){ 18 printf("0 "); 19 continue; 20 } 21 22 if(number==1){ 23 printf("1 "); 24 continue; 25 } 26 27 if(number==2){ 28 b%=4; 29 30 if(b==0) 31 b=4; 32 33 printf("%c ",two[b-1]); 34 continue; 35 } 36 37 if(number==3){ 38 b%=4; 39 40 if(b==0) 41 b=4; 42 43 printf("%c ",three[b-1]); 44 continue; 45 } 46 47 if(number==4){ 48 b%=2; 49 50 if(b==0) 51 b=2; 52 53 printf("%c ",four[b-1]); 54 continue; 55 } 56 57 if(number==5){ 58 printf("5 "); 59 continue; 60 } 61 62 if(number==6){ 63 printf("6 "); 64 continue; 65 } 66 67 if(number==7){ 68 b%=4; 69 70 if(b==0) 71 b=4; 72 73 printf("%c ",seven[b-1]); 74 continue; 75 } 76 77 if(number==8){ 78 b%=4; 79 80 if(b==0) 81 b=4; 82 83 printf("%c ",eight[b-1]); 84 continue; 85 } 86 87 if(number==9){ 88 b%=2; 89 90 if(b==0) 91 b=2; 92 93 printf("%c ",nine[b-1]); 94 continue; 95 96 } 97 98 } 99 100 return 0; 101 }