A^X mod P
Problem 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
1 <= n <= 10^6
0 <= A, K, a, b <= 10^9
1 <= m, P <= 10^9
Output
c is the case number start from 1.
ans is the answer of this problem.
Example Input
23 2 1 1 1 100 1003 15 123 2 3 1000 107
Example Output
Case #1: 14Case #2: 63
Hint
Author
题意:
题目意思很简单,就是求(A^f[1]+A^f[2]+。。。+A^f[n])%P
题解:
一开始直接扫描一遍结果无情TL,用快速幂计算幂值有很多重复的计算,因此想办法将结果保存在数组里面,dp的思想。显然f[i]=fix*k+j,这样分解是对的,那么选取一个合适的fix,这样数组可以存下需要的解。不妨令fix=31623
,那么A^(fix*k+j)%m,就可以分解成(A^k)^fix*A^j,用dpk[i]保存(A^k)^i,dpj[i]保存A^i,那么
A^f[i]=dpk[i/fix]*dpj[i%fix];
#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[0]=1; for(i=1;i<=maxn;i++){ X[i]=(X[i-1]*A)%P; } LL tmp=X[maxn]; Y[0]=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,&m,&P); Init(); Solve(icase); } }