upc.2219: A^X mod P(打表 && 超越快速幂（in some ways)）-优快云博客

本文介绍了一种解决ACM竞赛中涉及大量快速幂运算问题的方法。通过将指数拆分为小数部分和大数部分，并预处理两种情况下的结果，实现了在O(1)的时间复杂度内快速求解。适用于指数范围为1到10^9的问题。

2219: A^X mod P

Time Limit: 5 Sec Memory Limit: 128 MB Submit: 417 Solved: 68 [ Submit][ Status][ Web Board]

Description

It's easy for ACMer to calculate A^X mod P. Now given seven integers n, A, K, a, b, m, P, and a function f(x) which defined as following.

f(x) = K, x = 1

f(x) = (a*f(x-1) + b)%m , x > 1

Now, Your task is to calculate

( A^(f(1)) + A^(f(2)) + A^(f(3)) + ...... + A^(f(n)) ) modular P.

Input

In the first line there is an integer T (1 < T <= 40), which indicates the number of test cases, and then T test cases follow. A test case contains seven integers n, A, K, a, b, m, P in one line.

1 <= n <= 10^6

0 <= A, K, a, b <= 10^9

1 <= m, P <= 10^9

Output

For each case, the output format is “Case #c: ans”.

c is the case number start from 1.

ans is the answer of this problem.

Sample Input

2
3 2 1 1 1 100 100
3 15 123 2 3 1000 107

Sample Output

Case #1: 14
Case #2: 63

HINT

Source

2013年山东省第四届ACM大学生程序设计竞赛

 1 #include<stdio.h>
 2 typedef long long ll ;
 3 int T ;
 4 int  n, A, K, a, b, m, P ;
 5 int small [1 << 15 | 5] ;
 6 int big [1 << 15 | 5] ;
 7 int ans ;
 8 
 9 void solve ()
10 {
11     int ret = 0 ;
12     small[0] = 1 % P ;
13     for (int i = 1 ; i < (1 << 15 | 5) ; i++) {
14         small [i] = (ll) small[i - 1] * A % P ;
15     }
16     big[0] = 1 % P ;
17     for (int i = 1 ; i < (1 << 15 ) ; i++) {
18         big[i] = (ll) big[i - 1] * small [1 << 15] % P ;
19     }
20     while (n --) {
21         ret += (ll) small [K & (1 << 15) - 1] * big [K >> 15] % P ;
22         if (ret >= P) ret -= P ;
23         K = ((ll) a * K + b) % m ;
24     }
25     printf ("Case #%d: %d\n" , ++ans , ret ) ;
26 }
27 
28 int main ()
29 {
30     //freopen ("a.txt" , "r" , stdin );
31     scanf ("%d" , &T) ;
32     ans = 0 ;
33     while (T--) {
34         scanf ("%d%d%d%d%d%d%d" , &n , &A , &K , &a , &b , &m , &P) ;
35         solve () ;
36     }
37     return 0 ;
38 }