【杭电1061】Rightmost Digit 求n^n，找规律！！

 Rightmost Digit

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit Status

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.          

刚开始按传统方法，总是超时。。。不能暴力解决  比较大的不能用pow，结果为浮点型，精度不准确，只能保证前11位正确

超时代码：

#include<stdio.h>int f(int b){int i,t=1;for(i=1;i<=b;i++)t=t*b;return t;}int main(){int a,n;int t;while(scanf("%d",&n)!=EOF){while(n--){scanf("%d",&a);t=f(a)%10;printf("%d\n",t);}}return 0;}

正确代码（参考代码）：

可以发现一个规律：当   n = 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 27 28 29 30 31 ...rdigit = 1 4 7 6 5 6 3 6 9   0   1   6  3   6   5   6   7   4  9  0  1   4   7   6   5   6   3   6  9   0    ...所以是以20为周期的规律。这样就简单多了。。

12345678910111213
#include<stdio.h>   int  t[22]={0,1,4,7,6,5,6,3,6,9,0,1,6,3,6,5,6,7,4,9,0};  int main ()  {      int m,n;       scanf("%d",&m) ;      while(m--)      {        scanf("%d",&n);        printf("%d\n",t[n%20]) ;     }     return 0 ; }

下面是 m^n  % k 的快速幂：模板
// m^n % k
int quickpow(int m,int n,int k)
{
    int b = 1;
    while (n > 0)
    {
          if (n & 1)
             b = (b*m)%k;
          n = n >> 1 ;
          m = (m*m)%k;
    }
    return b;
} 
下面是套用的模板，在杭电上提交不知道为啥是wrong answer请高手指点！！\(^o^)/~

#include<stdio.h>int quickpow(int m,int n,int k){    int b = 1%k;    while (n > 0)    {          if (n & 1)             b = (b*m)%k;          n = n >> 1 ;          m = (m*m)%k;    }    return b;} int main(){int a,t,n;scanf("%d",&n);while(n--){scanf("%d",&a);t=quickpow(a,a,10);printf("%d\n",t);}return 0;}

下面也是套用模板，刚开始wrong answer why！！原来没有先对n,m取余，找了好久错误，哎。。。

#include<cstdio>int quickpow(int n,int m,int mod){int b=1%mod,t=n%mod;n,m较大要先取余

while(m){if(m&1){b=(t*b)%mod;//2^10=2^8*2^2=（（2^2）^2）^2*2^2；不断开出2；}base=(t*t)%mod;m>>=1;//右移一位，相当于除以2；}return b;}int main(){int a,n;scanf("%d",&n);while(n--){scanf("%d",&a);printf("%d\n",quickpow(a,a,10));}return 0;}