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http://acm.hdu.edu.cn/showproblem.php?pid=1061
Given a positive integer N, you should output the most right digit of N^N.
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).
For each test case, you should output the rightmost digit of N^N.
3
3
5
7
5
快速幂。。
#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<vector>
#include<map>
using std::map;
using std::min;
using std::find;
using std::pair;
using std::vector;
using std::multimap;
#define pb(e) push_back(e)
#define sz(c) (int)(c).size()
#define mp(a, b) make_pair(a, b)
#define all(c) (c).begin(), (c).end()
#define iter(c) __typeof((c).begin())
#define cls(arr, val) memset(arr, val, sizeof(arr))
#define cpresent(c, e) (find(all(c), (e)) != (c).end())
#define rep(i, n) for(int i = 0; i < (int)n; i++)
#define tr(c, i) for(iter(c) i = (c).begin(); i != (c).end(); ++i)
const int N = 100001;
const int INF = 0x3f3f3f3f;
typedef unsigned long long ull;
ull pow_mod(int n) {
ull ans = 1, x = n;
while(n) {
if(n & 1) ans = ans * x % 10;
x = x * x % 10;
n >>= 1;
}
return ans;
}
int main() {
#ifdef LOCAL
freopen("in.txt", "r", stdin);
freopen("out.txt", "w+", stdout);
#endif
int t, n;
scanf("%d", &t);
while(t--) {
scanf("%d", &n);
printf("%lld\n", pow_mod(n));
}
return 0;
}
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原文地址：http://www.cnblogs.com/GadyPu/p/4792673.html
