# Sdut 2605-a^x MoD P (large power decomposition summation)

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A^x MoD P Time limit:5000ms Memory limit:65536k have questions? Dot here ^_^ Title 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

(a^ (f (1)) + a^ (f (2)) + a^ (f (3)) + ... + a^ (f (n))) Modular P.

Enter the first line there are an integer T (1 < T <=), 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 format is ' case #c: ans ' for each case.

The case number of C is the start from 1.

Ans is the answer of this problem.

Sample input
`23 2 1 1 1 100 1003 15 123 2 3 1000 107`
Sample output
`Case #1:14Case #2:63`
Hint Source 2013 ACM University student Program Design competition in Shandong Province

Test instructions: a^x mod P's and.

PS: Due to kneel, with a quick power to definitely not, time out, and then use a quick power + fast multiplication or time-out, has been optimized optimization is also good, had to see the next puzzle, really to kneel.

Idea: This used the Decomposition method, the a^f in the F decomposition into I * k + j form. Save it in an array, just look it up when you use it.

`#include <stdio.h> #include <math.h> #include <string.h> #include <stdlib.h> #include < iostream> #include <sstream> #include <algorithm> #include <set> #include <queue> #include <stack> #include <map>using namespace std;typedef long long ll;const int inf=0x3f3f3f3f;const double pi= ACOs ( -1.0); const int maxn=33333; LL X[MAXN+10],Y[MAXN+10];    LL n,a,k,a,b,m,p;void Init () {int i;    X=1;    for (i=1;i<=maxn;i++) {x[i]= (x[i-1]*a)%P;    } LL TMP=X[MAXN];    Y=1;    for (i=1;i<=maxn;i++) {y[i]= (y[i-1]*tmp)%P;    }}void Solve (int icase) {int i;    LL fx=k;    LL res=0;        for (i=1;i<=n;i++) {res= (res+ (Y[FX/MAXN]*X[FX%MAXN])%P)%P;    fx= (a*fx+b)%m; } printf ("Case #%d:%lld\n", icase,res);}    int main () {int t,icase;    scanf ("%d", &t); for (icase=1;icase<=t;icase++) {scanf ("%lld%lld%lld%lld%lld%lld%lld", &n,&a,&k,&a,&b,&am        P;M,&AMP;P);   Init ();     Solve (icase); }}`

Sdut 2605-a^x MoD P (large power decomposition summation)

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