Problem 1097

最新推荐文章于 2022-10-17 08:27:38 发布

最新推荐文章于 2022-10-17 08:27:38 发布 · 218 阅读

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本文介绍了一种解决特定数学问题的算法：给定两个正整数a和b，如何高效地计算a^b的最后一位数字。通过使用余数的概念并限制计算范围来优化算法，避免了直接计算大数值的复杂性和资源消耗。

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

#include <iostream>

using namespace std;

int main()
{
 int a, b, result = 1, temp, has = 0, pos, cap = 0;
 int yushu[10];

 while (cin >> a >> b)
 {
 temp = b;
 has = 0;
 pos = b;
 cap = 0;
 for(int i = 0; i < 10; i++)
 {
 yushu[i] = 0;
 }
 while(a > 9)
 {
 a %= 10;
 }
 while(temp--)
 {
 result *= a;
 while(result > 9)
 {
 result %= 10;
 }
 for(int i = 0; i < 10; i++)
 {
 if(yushu[i] == result)
 {
 has = 1;
 break;
 }
 }
 if(has == 1)
 break;
 else
 {
 yushu[cap] = result;
 cap++;
 }
 }
 if(has == 1 && cap != 0)
 {
 pos = b % (cap) - 1;
 if(pos == -1)
 pos = cap - 1;
 cout << yushu[pos] << endl;
 }
 else
 if(has != 1)
 cout << result << endl;
 if(has == 1 && cap == 0)
 cout << 0 << endl;
 result = 1;
 a = 0;
 b = 0;
 for(int i = 0; i < 10; i++)
 {
 yushu[i] = 0;
 }
 }
 return 0;
}