A hard puzzle(HDU1097)

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本文探讨了如何快速计算a^b的最后一位数字，通过分析2到9的幂次方尾数规律，找到了一种简单的方法来解决这个问题，并给出了具体的实现代码。

A hard puzzle

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 22701 Accepted Submission(s): 7957

Problem Description

lcy gives a hard puzzle to feng5166,lwg,JGShining and Ignatius: gave a and b,how to know the a^b.everybody objects to this BT problem,so lcy makes the problem easier than begin.
this puzzle describes that: gave a and b,how to know the a^b's the last digit number.But everybody is too lazy to slove this problem,so they remit to you who is wise.

Input

There are mutiple test cases. Each test cases consists of two numbers a and b(0<a,b<=2^30)

Output

For each test case, you should output the a^b's last digit number.

Sample Input

7 66
8 800

Sample Output

9
6

解题思路：把2 ~9 的 N次方打表找规律；

打表：

#include <cstdio> 
#include <iostream> 
using namespace std;
 int main() 
{
       int i , j; 
       for(i = 2; i <= 9 ; i++) 
       { 
            cout <<"a = " << i <<endl;
            int k = i , sum = 1; 
            for(j = 1; j <= 16 ; j++) 
           { 
                       sum *= k; 
                      while(sum > 10) 
                      { 
                             sum %= 10; 
                       } 
                       cout << "b = " << j << "," << "a = " << sum <<'\t';
              } cout << endl; 
        } 
        return 0;
}
/*
尾数表：

a = 2
b = 1,a = 2     b = 2,a = 4     b = 3,a = 8     b = 4,a = 6     b = 5,a = 2
b = 6,a = 4     b = 7,a = 8     b = 8,a = 6     b = 9,a = 2     b = 10,a = 4
b = 11,a = 8    b = 12,a = 6    b = 13,a = 2    b = 14,a = 4    b = 15,a = 8
b = 16,a = 6
a = 3
b = 1,a = 3     b = 2,a = 9     b = 3,a = 7     b = 4,a = 1     b = 5,a = 3
b = 6,a = 9     b = 7,a = 7     b = 8,a = 1     b = 9,a = 3     b = 10,a = 9
b = 11,a = 7    b = 12,a = 1    b = 13,a = 3    b = 14,a = 9    b = 15,a = 7
b = 16,a = 1
a = 4
b = 1,a = 4     b = 2,a = 6     b = 3,a = 4     b = 4,a = 6     b = 5,a = 4
b = 6,a = 6     b = 7,a = 4     b = 8,a = 6     b = 9,a = 4     b = 10,a = 6
b = 11,a = 4    b = 12,a = 6    b = 13,a = 4    b = 14,a = 6    b = 15,a = 4
b = 16,a = 6
a = 5
b = 1,a = 5     b = 2,a = 5     b = 3,a = 5     b = 4,a = 5     b = 5,a = 5
b = 6,a = 5     b = 7,a = 5     b = 8,a = 5     b = 9,a = 5     b = 10,a = 5
b = 11,a = 5    b = 12,a = 5    b = 13,a = 5    b = 14,a = 5    b = 15,a = 5
b = 16,a = 5
a = 6
b = 1,a = 6     b = 2,a = 6     b = 3,a = 6     b = 4,a = 6     b = 5,a = 6
b = 6,a = 6     b = 7,a = 6     b = 8,a = 6     b = 9,a = 6     b = 10,a = 6
b = 11,a = 6    b = 12,a = 6    b = 13,a = 6    b = 14,a = 6    b = 15,a = 6
b = 16,a = 6
a = 7
b = 1,a = 7     b = 2,a = 9     b = 3,a = 3     b = 4,a = 1     b = 5,a = 7
b = 6,a = 9     b = 7,a = 3     b = 8,a = 1     b = 9,a = 7     b = 10,a = 9
b = 11,a = 3    b = 12,a = 1    b = 13,a = 7    b = 14,a = 9    b = 15,a = 3
b = 16,a = 1
a = 8
b = 1,a = 8     b = 2,a = 4     b = 3,a = 2     b = 4,a = 6     b = 5,a = 8
b = 6,a = 4     b = 7,a = 2     b = 8,a = 6     b = 9,a = 8     b = 10,a = 4
b = 11,a = 2    b = 12,a = 6    b = 13,a = 8    b = 14,a = 4    b = 15,a = 2
b = 16,a = 6
a = 9
b = 1,a = 9     b = 2,a = 1     b = 3,a = 9     b = 4,a = 1     b = 5,a = 9
b = 6,a = 1     b = 7,a = 9     b = 8,a = 1     b = 9,a = 9     b = 10,a = 1
b = 11,a = 9    b = 12,a = 1    b = 13,a = 9    b = 14,a = 1    b = 15,a = 9
b = 16,a = 1

从表格可以发现 ， 每乘 四次 ， a的最后一位数都会循环 ；
最终代码：*/
#include <iostream>
#include <cstdio>
using namespace std;
int main()
{
	int num[10][4] = {{0, 0 , 0 , 0},//用二维数组把 0-9 的个尾储存起来；
	              {1 ,1 ,1 , 1  },
	              {6 ,2 ,4 , 8 },
				  {1 ,3 ,9 , 7 },
				  {6 , 4 , 6 , 4},
				  {5 , 5 , 5 , 5},
				  {6 , 6 , 6 , 6},
				  {1 ,7 ,9 , 3 },
				  {6 ,8 ,4 , 2 },
				  {1 ,9 ,1 , 9 }};
	int a , b;
	while(scanf("%d%d" ,&a , &b) != EOF)
	{
		printf("%d\n" , num[a%10][b%4]);
	}
	return 0;
}