Rightmost Digit
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 16539 Accepted Submission(s): 6355
Problem Description
Given a positive integer N, you should output the most right digit of N^N.
Input
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).
Each test case contains a single positive integer N(1<=N<=1,000,000,000).
Output
For each test case, you should output the rightmost digit of N^N.
Sample Input
2 3 4
Sample Output
7 6HintIn the first case, 3 * 3 * 3 = 27, so the rightmost digit is 7. In the second case, 4 * 4 * 4 * 4 = 256, so the rightmost digit is 6.
Author
Ignatius.L
这是我用递归写的此题,代码如下:
#include <stdio.h>
int main()
{
int f(int n,int s);
int n,m,s,x,j;
scanf("%d",&m);
while(m--)
{
x=0;
scanf("%d",&n);
s=n%10;
j=f(n,s);
printf("%d\n",j);
}
return 0;
}
int f(int n,int s)
{
int j;
if(n==1)
{
return (s);
}else if(n==2)
{
return (s*s%10);
}else
{
if(n%2==0)
{
j=f(n/2,s);
j=j*j;
j=j%10;
}else
{
j=f(n/2,s);
j=j*j*s;
j=j%10;
}
}
return (j);
}