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HDU 1061 Rightmost Digit(分类讨论)

2014-12-07 18:16 337 查看

Rightmost Digit

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6437 Accepted Submission(s): 1673
[align=left]Problem Description[/align]
Given a positive integer N, you should output the most right digit of N^N.

[align=left]Input[/align]
The input contains several test cases. The first line of the input is a single integer T which is the number of test cases. T test cases follow.

Each test case contains a single positive integer N(1<=N<=1,000,000,000).

[align=left]Output[/align]
For each test case, you should output the rightmost digit of N^N.

[align=left]Sample Input[/align]

2
3
4


[align=left]Sample Output[/align]

7
6


#include<stdio.h>
int main()
{
int n,a,i,cases,s=1;
scanf("%d",&cases);
while(cases>0)
{
scanf("%d",&n);
a=n%10;
if(a==1||a==0||a==5||a==6)
printf("%d\n",a);
if(a==2)
switch(n%4)
{
case 1:printf("2\n");break;
case 2:printf("4\n");break;
case 3:printf("8\n");break;
case 0:printf("6\n");break;
}
if(a==3)
switch(n%4)
{
case 1:printf("3\n");break;
case 2:printf("9\n");break;
case 3:printf("7\n");break;
case 0:printf("1\n");break;
}
if(a==7)
switch(n%4)
{
case 1:printf("7\n");break;
case 2:printf("9\n");break;
case 3:printf("3\n");break;
case 0:printf("1\n");break;
}
if(a==8)
switch(n%4)
{
case 1:printf("8\n");break;
case 2:printf("4\n");break;
case 3:printf("2\n");break;
case 0:printf("6\n");break;
}
if(a==4)
switch(n%2)
{
case 1:printf("4\n");break;
case 0:printf("6\n");break;
}
if(a==9)
switch(n%2)
{
case 1:printf("9\n");break;
case 0:printf("1\n");break;
}
cases--;
}
return 0;
}
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