hdoj1061 Rightmost Digit（快速幂+简单思路）

Description

Given a positive integer N, youshould 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).

Output

For each test case, you should output the rightmost digit of N^N.

Sample Input

2

3

4

Sample Output

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.

题意：求n的n次方的最后一位数。

思路：快速幂，降幂计算不会超时。

代码：

#include<cstdio>
__int64 quickpow(__int64 n,__int64 m,__int64 mod)
{
__int64 ans=1,base=n;
while(m)
{
if(m&1)
{
ans=(base*ans)%mod;
}
base=(base*base)%mod;
m/=2;
}
return ans;
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
__int64 n;
scanf("%I64d",&n);
printf("%I64d\n",quickpow(n,n,10));
}
return 0;
}

还有一种方法就是用字符串。记录n的最后一位数，0～9的n次方尾数至多有四种情况。

0.1.5.6.9的n次方还是等于其本身

2=[2,4,8,6],3=[3,9,7,1],4=[4,6],7=[7,9,3,1],8=[8,4,2,6].

所以可以用n对4取余是几就是第几个数。可以打表来做也可以直接写

打表的话char a[4][10]={{0,1,2,3,4,5,6,7,8,9},{0,1,4,9,6,5,6,9,4,1},{0,1,8,7,4,5,6,3,2,9},{0,1,6,1,5,6,1,6,1}};

对四取余的m后直接输出a[m]
.

代码：

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string.h>
using namespace std;
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
int a;
scanf("%d",&a);
int k=a%10;
if(k==0||k==1||k==5||k==6||k==9)
{
printf("%d\n",k);
}
else
{
if(k==2)
{
if(a%4==0)
printf("6\n");
else
printf("4\n");
}
if(k==3)
{
if(a%4==1)
printf("3\n");
else
printf("7\n");
}
if(k==4)
printf("6\n");
if(k==7)
{
if(a%4==1)
printf("7\n");
else
printf("3\n");
}
if(k==8)
{
if(a%4==0)
printf("6\n");
else
printf("4\n");
}
}
}
return 0;
}