第四届山东省赛 A^X mod P [预处理]【思维】

来源：互联网发布：砸炮纸淘宝能买到吗编辑：程序博客网时间：2024/05/12 14:47

题目链接：http://acm.upc.edu.cn/problem.PHP?id=2219
————————————————————————————————————
2219: A^X mod P
Time Limit: 5 Sec Memory Limit: 128 MB
Submit: 142 Solved: 28
[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

————————————————————————————————————

题目大意：
就是给你7个数:
n,A,K,a,b,m,P
一个公式:
f(x)={ K(a∗f(x−1)+b)%m,x=1,x>1

问：
(Af(1)+Af(2)+Af(3)+......+Af(n))modP.

解题思路:
其实思路很好构建 ,
只要求出f(i)然后O(nlogn)的快速幂就能求解

但是这题卡了log ,所以不能快速幂

对于求一次幂用快速幂会非常快但是求多次就不是很快了

我们要先预处理出所有{Ax|x∈[1,P]}

发现根本存不下，P≤109，
但是我们可以预处理出
{Ax|x∈[1,P−−√]}
和
{(AP√)x|x∈[1,P−−√]}
这样我们就可以通过一次相乘,开快速求出Z∈{Ax|x∈[1,P]}了

这样下来复杂度就变成了O(Tn)

附本题代码

——————————————————————————————————————————————————————————————

#include<bits/stdc++.h>using namespace std;typedef long long int LL;/********************************/LL Ai[100005];LL A1e5i[100005];int main(){    int _ = 1,kcase = 0;    scanf("%d",&_);    while(_--){        int n, A, K, a, b, m, P;        scanf("%d%d%d%d%d%d%d",&n, &A, &K, &a, &b, &m, &P);        Ai[0]=1;        Ai[1]=A;        for(int i=1;i<=100000;i++){            Ai[i]=Ai[i-1]*A%P;        }        A1e5i[0]=1;        A1e5i[1]=Ai[100000];        for(int i=2;i<=100000;i++){            A1e5i[i]=A1e5i[1]*A1e5i[i-1]%P;        }        LL ans =0,tmp=K;        for(int i=1;i<=n;i++){            ans+=Ai[tmp%100000]*A1e5i[tmp/100000];            (ans>P)?ans%=P:true;            tmp=tmp*a+b,tmp%=m;        }        printf("Case #%d: ",++kcase);        printf("%lld\n",ans);    }    return 0;}

0 0