HDU 1061 Rightmost Digit

链接：http://acm.hdu.edu.cn/showproblem.php?pid=1061

Rightmost Digit

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

Total Submission(s): 39835    Accepted Submission(s): 15036

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.

Author

Ignatius.L

Recommend

大意——给你一个正整数n，要你算出n^n的最后一位数码。其中1<=n<=1,000,000,000。

思路——与HDU 1097差不多，我们可以用二分求快速幂，并且边乘边模，这样的话就不会出问题了。

复杂度分析——时间复杂度:O(log(n)),空间复杂度:O(1)

附上AC代码：

#include <iostream>#include <cstdio>#include <string>#include <cmath>#include <iomanip>#include <ctime>#include <climits>#include <cstdlib>#include <cstring>#include <algorithm>#include <queue>#include <vector>#include <set>#include <map>using namespace std;typedef unsigned int UI;typedef long long LL;typedef unsigned long long ULL;typedef long double LD;const double pi = acos(-1.0);const double e = exp(1.0);const double eps = 1e-8;const int mod = 10;int quick_pow(int a);int main(){ios::sync_with_stdio(false);int T, a;scanf("%d", &T);while (T--){scanf("%d", &a);printf("%d\n", quick_pow(a));}return 0;}int quick_pow(int a){int res = 1;int b = a;while (b > 0){if (b & 1)res = ((res%mod)*(a%mod))%mod;a = ((a%mod)*(a%mod))%mod;b >>= 1;}return res;}