HDOJ–1061–Rightmost Digit

2018年04月24日 ⁄ 综合 ⁄ 共 978字 ⁄ 字号 评论关闭
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.
```

Author
Ignatius.L

```#include<stdio.h>
#include<cstring>
#include<algorithm>
#include<math.h>
#include<iostream>
#include<time.h>
using namespace std;
char a[10000];
int main()
{
long long int n;
int i,m,q;
int l,t,j,x,l2,b[50];
scanf("%d",&t);
while(t--)
{
m=0;
scanf("%lld",&n);
sprintf(a,"%lld",n);
l=strlen(a);
a[l-1]=a[l-1]-48;
x=a[l-1];
b[0]=x;
for(i=1;;i++)
{
b[i]=b[i-1]*x;
if(b[i]>=10)
b[i]=b[i]%10;
for(j=0;j<i;j++)
if(b[i]==b[j])
{
m=1;
q=i;
}
if(m)
break;
}
x=n%q;
if(x-1>=0)
x=x-1;
else
x=q-1;
printf("%d\n",b[x]);
}
return 0;
}```