A. Arpa’s hard exam and Mehrdad’s naive cheat
2016-12-07 21:54
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time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
There exists an island called Arpa’s land, some beautiful girls live there, as ugly ones do.
Mehrdad wants to become minister of Arpa’s land. Arpa has prepared an exam. Exam has only one question, given n, print the last digit
of 1378n.
Mehrdad has become quite confused and wants you to help him. Please help, although it's a naive cheat.
Input
The single line of input contains one integer n (0 ≤ n ≤ 109).
Output
Print single integer — the last digit of 1378n.
Examples
input
output
input
output
Note
In the first example, last digit of 13781 = 1378 is 8.
In the second example, last digit of 13782 = 1378·1378 = 1898884 is 4.
解题说明:水题,找规律发现末尾最后数字是一个循环。
#include<cstdio>
#include <cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<vector>
#include <map>
using namespace std;
int a[5]={6,8,4,2,6};
int main()
{
int n,i;
scanf("%d", &n);
if (n==0)
{
printf("1\n");
}
else
{
printf("%d\n",a[n%4]);
}
return 0;
}
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
There exists an island called Arpa’s land, some beautiful girls live there, as ugly ones do.
Mehrdad wants to become minister of Arpa’s land. Arpa has prepared an exam. Exam has only one question, given n, print the last digit
of 1378n.
Mehrdad has become quite confused and wants you to help him. Please help, although it's a naive cheat.
Input
The single line of input contains one integer n (0 ≤ n ≤ 109).
Output
Print single integer — the last digit of 1378n.
Examples
input
1
output
8
input
2
output
4
Note
In the first example, last digit of 13781 = 1378 is 8.
In the second example, last digit of 13782 = 1378·1378 = 1898884 is 4.
解题说明:水题,找规律发现末尾最后数字是一个循环。
#include<cstdio>
#include <cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<vector>
#include <map>
using namespace std;
int a[5]={6,8,4,2,6};
int main()
{
int n,i;
scanf("%d", &n);
if (n==0)
{
printf("1\n");
}
else
{
printf("%d\n",a[n%4]);
}
return 0;
}
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