Pupu（hdu3003）数论

Pupu

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 913    Accepted Submission(s): 385

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

2

3

0

 

Sample Output

1

2

 

o(︶︿︶)o 唉，好不容易想到方法的，，，居然超时了，

 

思路：我们可以将此题用二进制的思想来解题。0代表不透明，1代表透明。

2层： （0 0）->（1 0）->（0 1）三天

3层：（0 0 0）->（1 0 0)-> (0 1 0)->(1 1 0)->(0 0 1) 五天

所以题意就转变为求2的n-1次方%n。

可是刚刚开始做的时候超时了，要对2的n-1次方这个步骤进行优化，就过了- -||

ps:http://acm.hdu.edu.cn/showproblem.php?pid=3003

 

/*超时。。。。呜呜#include <stdio.h>int main(){    __int64 n,m,i,j,a;    while(scanf("%I64d",&n),n)    {        m=n-1;        a=2;        for(i=1;i<m;)        {            if(i*2<=m)            {                a=a*a%n;                i=i*2;            }            else            {                a=a*2%n;                i++;            }        }        printf("%I64d\n",(a+1)%n);    }    return 0;}*/

优化后的代码。。。

#include<stdio.h>int main(){    __int64 n,m,i,k,a,bit[100];    while(scanf("%I64d",&n),n)    {        a=1;        k=0;        m=n-1;        while(m)        {            bit[k++]=m%2;//判断奇偶            m=m>>1;//m/=2;        }        for(i=k-1;i>=0;i--)        {            a=a*a%n;            if(bit[i]==1)                a=a*2%n;        }        printf("%d\n",(a+1)%n);    }    return 0;}