A^B%C 要用C++写 我只会C++

[tsshouhu]

(A^B)%C

Time Limit:1000MS Memory Limit:65536K
Total Submit:263 Accepted:35

Description

Naruto' faith is never give up, so I hope you can persist in ACM. ACM is not a shortcut to success, there are many obstacles you should overcome, but only overcome them you can become more powerful !
First, you need to overcome English. Now this problem.


Input

The input consist of three integers A,B,C; 1<= A <=10000; 1<= B <=10^9; 1<=C<=10000; three 0 sign the end of the input.

Output

The output will be a single integer——(A^B)%C. For example,A=2,B=4,C=5, then the result is (2^4)%5 = (2*2*2*2)%5 = 1.

Sample Input


2 2 3
3 3 4
4389 99999999 1303
0 0 0


Sample Output


1
3
698

[卡卡篷]

前两个对上了，第三个没对上，代码应该没有问题才对
试了很多数据，LZ看下你提供的数据和答案都没问题吗？

#include <iostream>
using namespace std:

typedef long unsigned int LUINT;

LUINT remainder(LUINT a, LUINT b, LUINT c)
{
	a %= c;
	bool bEven = (0==b%2);
	LUINT tmp = a;
	for (LUINT i=bEven?b:b-1; i>=2; i/=2)
	{
		a = (a*a % c);
	}
	return bEven? a%c : (a*tmp)%c;
}

int main()
{
	cout<<remainder(2, 2, 3)<<endl;
	cout<<remainder(3, 3, 4)<<endl;
	cout<<remainder(4389, 99999999, 1303)<<endl;
	
	return 0;
}

[卡卡篷]

上面的错了，下面的经测式正确

#include <iostream>
using namespace std;

typedef long unsigned int LUINT;

LUINT remainder(LUINT a, LUINT b, LUINT c)
{
	a %= c;
	if (!a)
	{
		return 0;
	}

	LUINT tmp = 1;
	for (LUINT i=b; i>=2; i/=2)
	{
		if (i%2)
		{
			tmp = tmp*a%c;
			--i;
		}
		a = a*a%c;
	}

	return a*tmp%c;
}

int main()
{	
	cout<<remainder(2, 2, 3)<<endl;
	cout<<remainder(3, 3, 4)<<endl;
	cout<<remainder(4389, 99999999, 1303)<<endl;	

	return 0;
}

[arong1234]

这种ACM之类的练习题都不自己做，搞ACM还有什么意义啊:)
而且，就这么简单的题目都不会，能说会C＋＋么？

[sinosinux]

[Quote=引用 3 楼 arong1234 的回复:]
这种ACM之类的练习题都不自己做，搞ACM还有什么意义啊:)
而且，就这么简单的题目都不会，能说会C＋＋么？
[/Quote]

同意