hdu---1061 Rightmost Digit

Rightmost Digit

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 35487 Accepted Submission(s): 13456



[align=left]Problem Description[/align]
Given a positive integer N, you should output the most right digit of N^N.

[align=left]Input[/align]
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).

[align=left]Output[/align]
For each test case, you should output the rightmost digit of N^N.

[align=left]Sample Input[/align]

2
3
4

[align=left]Sample Output[/align]

7
6

简单的找循环节

注意中间会爆int！！！

#include <iostream>
#include <algorithm>
#include <string.h>
#include <math.h>
#include <stdio.h>
#include <queue>
using namespace std;
#define N 10010
#define Mod 10
#define LL long long
/*

*/
int f[100010];
int main(){
//freopen("in.in","r",stdin);
//freopen("out.out","w",stdout);
int T;
while(scanf("%d",&T)!=EOF){
while(T--)
{
int n;scanf("%d",&n);
//找循环节
int t = n%Mod;
int num=1;
LL a = n;
f[1] = t;
while(1){
a = (a*n)%Mod;
num++;
f[num] = a;
if(a == t) break;
}
num--; //循环节为num
//循环节后的余数
f[0] = f[num];
printf("%d\n",f[n%num]);
}
}
return 0;
}