HDU，hdukmp
Pupu
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1133 Accepted Submission(s): 445
Problem DescriptionThere 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?
InputThere are many testcase, each testcase only contains one integer N, the number of skins, process until N equals 0
OutputMaybe 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
Source2009 Multi-University Training Contest 11 - Host by HRBEU
題意：PuPu有n層皮膚，每層皮膚都有2種狀態：透明和不透明，每層皮膚如果能被太陽照射到，則被太陽照射一天后都會變換狀態。
PuPu在出生的時候，所有的皮膚都為不透明的，直到每一層的皮膚都有過變為透明狀態的時候，PuPu也就長大了。
問有n層皮膚的PuPu幾天后能長大。
思路：可以推出公式 ans = （2^(n-1) + 1）%n , 然後按照快速冪模數算出來即可（沒用LL又WA了一次!）
AC代碼：
#include <cstdio>#include <cstring>#include <cstdlib>#include <algorithm>#include <cmath>#define LL long longusing namespace std;LL qmod(LL n){LL q = 2, mo = n, ans = 1;n--;while(n){if(n&1) ans = (ans * q) % mo;q = (q%mo*q%mo)%mo;n>>=1;}return (ans+1)%mo;}int main(){LL n;while(scanf("%I64d", &n), n){printf("%I64d\n", qmod(n));}return 0;}
