HDU/HDOJ 3609 Up-up 2010多校联合17场ZSTU

 

Up-up

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 771    Accepted Submission(s): 212

Problem Description

The Up-up of a number a by a positive integer b, denoted by a↑↑b, is recursively defined by:
a↑↑1 = a,
a↑↑(k+1) = a ^(a↑↑k)
Thus we have e.g. 3↑↑2 = 3 ³ = 27, hence 3↑↑3 = 3 ²⁷= 7625597484987 and 3↑↑4 is roughly 10 ^{3.6383346400240996*10^12}
The problem is give you a pair of a and k,you must calculate a↑↑k ,the result may be large you can output the answer mod 100000000 instead

 

Input

A pair of a and k .a is a positive integer and fit in __int64 and 1<=k<=200

 

Output

a↑↑k mod 100000000

 

Sample Input

   
   
   
   

    
    
    
    3 2
3 3
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    27
97484987
   
   
   
   

 

Source

2010 ACM-ICPC Multi-University Training Contest（17）——Host by ZSTU

 

这道题我WA了好久的说。。

先开始是5 2

10 3这两组数据过不了。。

后来改了之后还是一直WA。。

WA了10+之后我看了下discuss发现那个a和k居然可以都为0.。。。Orz。。我当时就崩溃了

不过题目也太不厚道了，都明明说了是正整数，居然还有0的情况。。。

 

这个题方法很简答

就是基于一个理论

a^b%c=a^(b%phi(c))%c

 

然后我是非递归写法和网上的主流写法貌似不一样。不过我认为要易懂一点

 

我的代码：

#include<stdio.h>

typedef __int64 ll;
ll mod=100000000;
ll m[205];

ll eular(ll n)
{
	ll i,ret=1;
	for(i=2;i*i<=n;i++)
	{
		if(n%i==0)
		{
			n=n/i;
			ret=ret*(i-1);
			while(n%i==0)
			{
				n=n/i;
				ret=ret*i;
			}
		}
		if(n==1)
			break;
	}
	if(n>1)
		ret=ret*(n-1);
	return ret;
}

void init()
{
	ll i;
	m[1]=mod;
	for(i=2;i<=204;i++)
		m[i]=eular(m[i-1]);
}

ll power(ll a,ll b,ll c)
{
	ll res=1;
    while(b)
    {
		if(b%2==1)  
			res=res*a%c;
		a=(a%c)*(a%c)%c;
		b=b/2;
    }
    return res;
}

int main()
{
	ll a,k,i,ans;
	init();
	while(scanf("%I64d%I64d",&a,&k)!=EOF)
	{
		ans=a;
		if(a==0&&k%2==0)
		{
			printf("1\n");
			continue;
		}
		for(i=k;i>=2;i--)
		{
			ans=ans%m[i];
			if(ans==0)
				ans=m[i];
			ans=power(a,ans,m[i-1]);
		}
		printf("%I64d\n",ans%mod);
	}
	return 0;
}