Codeforces Round #383 (Div. 2) 742A Arpa’s hard exam and Mehrdad’s naive cheat
2017-07-18 17:40
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A. Arpa’s hard exam and Mehrdad’s naive cheat
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.
题意:让你求出1378^n 的个位数是几
思路:其实就是求出8^n 的个位数是几
稍微列个表,就能发现:个位数满足:8,4,2,6
知道这个规律就十分简单了
#include<bits/stdc++.h>
using namespace std;
int main()
{
int num[5]={6,8,4,2};
int n;
while(~scanf("%d",&n))
{
if(n==0)
{
printf("1\n");
}
else
{
printf("%d\n",num[n%4]);
}
}
return 0;
}
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
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.
题意:让你求出1378^n 的个位数是几
思路:其实就是求出8^n 的个位数是几
稍微列个表,就能发现:个位数满足:8,4,2,6
知道这个规律就十分简单了
#include<bits/stdc++.h>
using namespace std;
int main()
{
int num[5]={6,8,4,2};
int n;
while(~scanf("%d",&n))
{
if(n==0)
{
printf("1\n");
}
else
{
printf("%d\n",num[n%4]);
}
}
return 0;
}
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