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



[align=left]Problem Description[/align]
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 103.6383346400240996*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

[align=left]Input[/align]
A pair of a and k .a is a positive integer and fit in __int64 and 1<=k<=200

[align=left]Output[/align]
a↑↑k mod 100000000

[align=left]Sample Input[/align]

3 2
3 3

[align=left]Sample Output[/align]

27
97484987

[align=left]Source[/align]
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;
}