HD-ACM算法专攻系列（2）——Rightmost Digit

题目描述：

源码：

/**/
#include"iostream"
using namespace std;
 
int main()
{
	int t, mod;
	long long n;
	cin>>t;
	for(int i = 0; i < t; i++)
	{
		cin>>n;
		mod = n % 10;
		if(mod == 0 || mod == 1 || mod == 5 || mod == 6)
		{
			cout<<mod<<endl;
		}
		else
		{
			switch(mod)
			{
				case 2:
					mod = n % 4;
					if(mod == 0)
					{
						cout<<6<<endl;
					}
					else if(mod == 1)
					{
						cout<<2<<endl;
					}
					else if(mod == 2)
					{
						cout<<4<<endl;
					}
					else
					{
						cout<<8<<endl;
					}
					break;
				case 3:
					mod = n % 4;
					if(mod == 0)
					{
						cout<<1<<endl;
					}
					else if(mod == 1)
					{
						cout<<3<<endl;
					}
					else if(mod == 2)
					{
						cout<<9<<endl;
					}
					else
					{
						cout<<7<<endl;
					}
					break;
				case 4:
					if(n % 2 == 1)
					{
						cout<<4<<endl;
					}
					else
					{
						cout<<6<<endl;
					}
					break;
				case 7:
					mod = n % 4;
					if(mod == 0)
					{
						cout<<1<<endl;
					}
					else if(mod == 1)
					{
						cout<<7<<endl;
					}
					else if(mod == 2)
					{
						cout<<9<<endl;
					}
					else
					{
						cout<<3<<endl;
					}
					break;
				case 8:
					mod = n % 4;
					if(mod == 0)
					{
						cout<<6<<endl;
					}
					else if(mod == 1)
					{
						cout<<8<<endl;
					}
					else if(mod == 2)
					{
						cout<<4<<endl;
					}
					else
					{
						cout<<2<<endl;
					}
					break;
				case 9:
					if(n % 2 == 1)
					{
						cout<<9<<endl;
					}
					else
					{
						cout<<1<<endl;
					}
					break;
			}
		}
	}
    return 0;
}