作为史上最强的刷子之一,zhx的老师让他给学弟(mei)们出n道题。zhx认为第i道题的难度就是i。他想要让这些题目排列起来很漂亮。
zhx认为一个漂亮的序列{ai}下列两个条件均需满足。
1:a1..ai是单调递减或者单调递增的。
2:ai..an是单调递减或者单调递增的。
他想你告诉他有多少种排列是漂亮的。因为答案很大,所以只需要输出答案模p之后的值。
Input Multiply test cases(less than 10001000). Seek EOF as the end of the file.
Output For each case, there are two integers nn and pp separated by a space in a line. (1≤n,p≤10181≤n,p≤1018)OutputFor each test case, output a single line indicating the answer.
Sample Input
2 233 3 5
Sample Output
2 1Hint
In the first case, both sequence {1, 2} and {2, 1} are legal. In the second case, sequence {1, 2, 3}, {1, 3, 2}, {2, 1, 3}, {2, 3, 1}, {3, 1, 2}, {3, 2, 1} are legal, so the answer is 6 mod 5 = 1 思路:最终推导式2^n-2 因为乘数可能超long,所以用到了快速乘法代码:
import java.util.Scanner; public class Main { public static long multi(long a,long b,long p){//快速乘法 long ans=0; while(b>0){ if((b&1)==1) ans=(ans+a)%p; a=(a+a)%p; b>>=1; } return ans; } public static long quick_pow(long a,long b,long p){//快速幂 long ans=1; while(b>0){ if((b&1)==1) ans=multi(ans,a,p)%p; a=multi(a,a,p)%p; b>>=1; } return ans; } public static void main(String[] args) { Scanner scan=new Scanner(System.in); while(scan.hasNext()){ long n=scan.nextLong(); long p=scan.nextLong(); if(p==1) System.out.println("0");//情况特判 else if(n==1) System.out.println("1"); else{ long ans=quick_pow(2,n,p)-2; if(ans<0) ans+=p;//注意ans为负数情况,因为快速幂计算取余-2可能小于0 System.out.println(ans); } } } }
HDU - 5187 zhx's contest(快速幂+快速乘法)
原文地址:https://www.cnblogs.com/qdu-lkc/p/12193763.html
时间: 2024-10-25 20:09:07