HDOJ Pupu 3003【快速幂】

Pupu

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1409    Accepted Submission(s): 539

Problem Description

There is an island called PiLiPaLa.In the island there is a wild animal living in it, and you can call them PuPu. PuPu is a kind of special animal, infant PuPus play under the sunshine, and adult PuPus hunt near the seaside. They fell happy every day.

But there is a question, when does an infant PuPu become an adult PuPu?
Aha, we already said, PuPu is a special animal. There are several skins wraping PuPu's body, and PuPu's skins are special also, they have two states, clarity and opacity. The opacity skin will become clarity skin if it absorbs sunlight a whole day, and sunshine can pass through the clarity skin and shine the inside skin; The clarity skin will become opacity, if it absorbs sunlight a whole day, and opacity skin will keep sunshine out.

when an infant PuPu was born, all of its skins were opacity, and since the day that all of a PuPu's skins has been changed from opacity to clarity, PuPu is an adult PuPu.

For example, a PuPu who has only 3 skins will become an adult PuPu after it born 5 days(What a pity! The little guy will sustain the pressure from life only 5 days old)

Now give you the number of skins belongs to a new-laid PuPu, tell me how many days later it will become an adult PuPu?

 

Input

There are many testcase, each testcase only contains one integer N, the number of skins, process until N equals 0

 

Output

Maybe an infant PuPu with 20 skins need a million days to become an adult PuPu, so you should output the result mod N

 

Sample Input

230

 

Sample Output

12

 

Source

2009 Multi-University Training Contest 11 - Host by HRBEU

 

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题目大意:
一种动物有n层皮肤，每层皮肤都有两种状态，透明或者不透明，透明的皮肤光照一天变为不透明，不透明的皮肤光照一天变为透明
光可以透过透明的皮肤照射到下一层。 刚出生的动物所有层的皮肤都是不透明的，每层皮肤都变过透明，则这只动物长大成人。(特别注意是变过，，而不是所有成都一起变透明)
问，给定一个N，表示N层皮肤，问第几天可以长大成人，得数MOD上N、
题解：
可以先推前几个数。得出2,3,5,9.。。可以推算出公式 2^(n-1) + 1 结果 mod n  就是一个快速幂

#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;typedef long long ll;ll fun(ll x,ll n,ll modn){ll res=1;while(n>0){if(n&1) res=res*x%modn;x=x*x%modn;n>>=1;}return res;}int main(){int n;while(scanf("%d",&n),n){printf("%lld\n",fun(2,(ll)(n-1),n)+1);}return 0;}