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 = 3³ = 27, hence 3↑↑3 = 3²⁷= 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 23 3

 

Sample Output

2797484987

 

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;}