hdoj 1061 Rightmost Digit

Rightmost Digit

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 47546    Accepted Submission(s): 17962


[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

Hint
In 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.

 1--9逐个列举的  有规律，，，，

#include<cstdio>
int main()
{
int t,r;
long long n;
scanf("%d",&t);
while(t--)
{
scanf("%lld",&n);
r=n%10;
if(r==0||r==1||r==5||r==6)
printf("%d\n",r);
else if(r==2)
{
if(n%4==0)	printf("6\n");
else if(n%4==1)	printf("2\n");
else if(n%4==2)	printf("4\n");
else printf("8\n");
}
else if(r==3)
{
if(n%4==0)	printf("1\n");
else if(n%4==1)	printf("3\n");
else if(n%4==2)	printf("9\n");
else	printf("7\n");
}
else if(r==4)
{
if(n%2==0)	printf("6\n");
else printf("4\n");
}
else if(r==7)
{
if(n%4==0)	printf("1\n");
else if(n%4==1)	printf("7\n");
else if(n%4==2)	printf("9\n");
else printf("3\n");
}
else if(r==8)
{
if(n%4==0)	printf("6\n");
else if(n%4==1)	printf("8\n");
else if(n%4==2)	printf("4\n");
else printf("2\n");
}
else
{
if(n%2==0)	printf("1\n");
else	printf("9\n");
}

}
return 0;
}

快速幂运算：

注意用 long long!!

#include<cstdio>
#include<cmath>
#define LL long long
LL mod_pow(LL x,LL n,LL mod)
{
LL ans=1;
while(n)
{
if(n&1)
{
ans=(ans*x)%mod;
}
x=(x*x)%mod;
n>>=1;
}
return ans;
}
int main()
{
LL n;
int t;
scanf("%d",&t);
while(t--)
{
scanf("%lld",&n);
printf("%lld\n",mod_pow(n,n,10));
}
return 0;
}