?pid=1061
Rightmost DigitTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 39858Accepted Submission(s): 15045
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.
题意：简单的快速幂题目。求a的b次方所得结果的最后一位数。
解题思路：这是一题较简单的快速幂。以下是我对快速幂的理解，具体看代码。
快速幂思路： 求a^b，是不是要把a乘b次呢？这样也行， 只是慢了点 例如求a^8，先把a乘a一次，得到a^2，然后把 a^2和a^2乘一次，得到a^4，然后把a^4和a^4乘一 下就得到a^8了 假如要求a^10呢？10=8+2，a^10=a^8*a^2 把10化为二进制， 10=1010(2),1010(2)=1000(2)+10(2)
类似的题型有：
<span style="font-size:18px;"><span style="font-size:24px;">#include<iostream>#include<cstdio>#include<cstring>#include<cmath>using namespace std;int q_pow(int a,int b){int ans=1;//ans初始化为1while(b>0){if(b&1) ans =(ans*a)%10;//如果b为奇数，把ans乘于aa =(a*a)%10;//把a变成a^a,并且%10,防止溢出b>>=1;//b右移一位，相当于除于2}return ans;}int main(){ int n,cas; cin>>n; while(n–) {cin>>cas;printf("%d\n",q_pow(cas%10,cas));//a%10在这是为了提速，原因自己想想 } return 0;}</span></span>
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那段岁月，无论从何种角度读你，你都完美无缺，
