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).
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 6HintIn 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
思路:不用高精度,只需以每一部得出的个位数继续相乘,最后取个位数即可。
#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; }