ACM水题系列 HDOJ 1061

Rightmost Digit

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 6526 Accepted Submission(s): 1688 

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

234

 

Sample Output

76
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.
 

题意，求N^N最右边的数。

思想：因为相同数相乘的个位数是有周期性规律的，周期不大于10，我们可以先求到N相乘的周期T，同时记录下不同的个位数存在数组r[]里，然后再用（N-1）%T得到该数的下标。

#include <iostream>#include <string.h>using namespace std;int main(){    int t;    int r[10];    bool l[10];    cin>>t;    while(t--){        memset(r,0,sizeof(r));        memset(l,0,sizeof(l));        int a,b,n=1;        unsigned long m;        cin>>m;        a=m%10;        b=a;        r[0]=b;        l[b]=true;        b = (b*a)%10;        while(!l[b]){            l[b] = true;            r[n++]=b;            b = (b*a)%10;        }        cout<<r[(m-1)%n]<<endl;    }    return 0;}