原题点击获取,嘻嘻
要求:Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
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).
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.
Author
Ignatius.L
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问题简述:
一开始依旧是记得ACM几乎必备的循环输入!!(就是别忘了输入例子数后,要实现输入足够例子后能够结束程序)
回到正题!!乍一看的字面意思:输入一个N值,计算N的N次方,然后把计算得出的这个数的个位数字输出(其实怎么可能出一道这么简单的题目…),即使把类型改为long long ,程序也应该会超时的。错了很多次后第一次我是找到规律来做出的这道题,再看发现的字面意思:那个N的范围你再看看,你再想想计算N的N次方是得到的值,是不是超大的!!(N可以用int类型,可是N的N次方的值用int怕会不够),所以这不是简单的求幂题!!于是考完周试后知道原来是用快速幂…具体可以参考一下烟波煮雨的博客,如果不理解当然也可以记下来,嘻嘻嘻,不过理解至上。
通过后期的努力VJ通过的快速幂求法的代码:
#include<iostream>
using namespace std;
long long poww(long long a, long long b, long long c)
{
long long ans = 1;
a = a % c;
while (b > 0)
{
if (b % 2 == 1) ans = (ans*a) % c;
b = b / 2;
a = (a*a) % c;
}
return ans;
}
int main()
{
long long a;
int T;
cin >> T;
while (T--)
{
scanf_s("%lld", &a);
int n=a % 10;
printf("%lld\n", poww(n, a,10));
}
return 0;
}
如果还是不懂快速幂也没有关系,可以看看Karen_Yu_的博客。
上面我有提到我一开始是找规律的方法的做出了赛题,其实在计算本上计一下,会发现这个尾数在20以内循环出现,{0,1,4,7,6,5,6,3,6,9,0,1,6,3,6,5,6,7,4,9},就是这个20个数字会不断的出现。
VS通过的找规律的代码:
#include <iostream>
#include<math.h>
using namespace std;
int main()
{
int a[100] = { 0,1,4,7,6,5,6,3,6,9,0,1,6,3,6,5,6,7,4,9 };
int T, N;
cin >> T;
while (T--&&cin>>N)
{
int j, p;
j = N % 20;
p = a[j];
printf("%d\n", p);
}
}