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.
我写的错误？？？错误的原因是超时，
//rightmost digit
#include<iostream>
using namespace std;
int main()
{
int n,tmp,m;
cin>>n;
while(n--){
cin>>m;
tmp=;
m%=;
// cout<<"m-==="<<m<<endl;
for(int i=;i<m;i++)
tmp=tmp*m%;
cout<<tmp<<endl;
}
}
正确代码：
#include<iostream>
using namespace std;
int main()
{
int a,ans,n=;
double dval = ;
int count=;
cin>>n;
while(n--)
{
cin>>a;
ans=;
count=a;
a = a%;
while (count)
{
if (count&==)
ans=(ans*a)%;
a=(a*a)%;
count>>=;
}
cout<<ans<<endl;
}
return ;
}
解法二：
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 % ;
if (m == )
m = ;
n = n % ;
cout <<int( pow(n, m)) % << endl;
}
return ;
}
PS:
//利用二进制输出 输入的数字
#include<iostream>
using namespace std;
int main()
{
int test,t,n,ans;
while(cin>>n){ test=n;
t=;
ans=;
while(test){
if(test&==) ans+=t;
t*=;
test>>=;
}
cout<<ans<<endl;
}
}