杭电 1061
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Rightmost Digit
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 39292 Accepted Submission(s): 14819
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
234
Sample Output
76
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.
分析:
读完题首先想到的是大数,但是这大数貌似也太大了,就算能放下,这么多大数乘法铁定超时,考虑优化,写个小程序打表观察能发现这样一个规律,n的次方是有周期性的,且周期为4,这样就好办了,对于给定的N,求N*N的个位数,只需算N的个位数的N%4次方,然后对10取余就是所要求的结果了。
#include<iostream>
#include<math.h>
using namespace std;
int main()
{
int t;
cin >> t;
while (t--)
{
int n, m;
cin >> n;
m = n % 4;
if (m == 0)
m = 4;
n = n % 10;
cout <<int( pow(n, m)) % 10 << endl;
}
return 0;
}
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