sgu 181 X-Sequence

181. X-Sequence

time limit per test: 0.5 sec.
memory limit per test: 4096 KB

input: standard
output: standard

Let {x_i} be the infinite sequence of integers:
1) x₀ = A;
2) x_i = (alpha * x_i-1^2 + beta * x_i-1 + gamma) mod M, for i >= 1.
Your task is to find x_k if you know A, alpha, beta, gamma, M and k.

Input

Given A (1 <= A <= 10000), alpha (0 <= alpha <= 100), beta (0 <= beta <= 100), gamma (0 <= gamma <= 100), M (1 <= M <= 1000), k (0 <= k <= 10^9). All numbers are integer.

Output

Write x_k.

Sample test(s)

Input

1 1 1 1 10 1

Output

3

因为M比较小，且此递推关系是只和上一状态相关，所以我们可以知道此必定有周期，我们可以暴力找到周期和周期开始的点，然后就可以轻松解决，本题注意K=0的情况，具体的我在代码中打了注释。

贴上代码：

#include<iostream>#include<cstring>#include<cstdio>#include<set>#include<algorithm>#include<vector>#include<cstdlib>#define inf 0xfffffff#define CLR(a,b) memset((a),(b),sizeof((a)))#define FOR(a,b) for(int a=1;a<=(b);(a)++)using namespace std;int const nMax = 1010;int const base = 10;typedef int LL;typedef pair<LL,LL> pij;//    std::ios::sync_with_stdio(false);int a,alpha,beta,gamma,k,M,b,A,T;int at[2*nMax];int ans[2*nMax];int main(){    scanf("%d%d%d%d%d%d",&a,&alpha,&beta,&gamma,&M,&k);    if(k==0) {       //注意K=0的情况，输出原值，不是%M的值        printf("%d\n",a);        return 0;    }    CLR(at,-1);    a%=M;    at[a]=0;    b=a;    ans[0]=a;    alpha%=M,beta%=M,gamma%=M;    for(int i=1;;i++){        a=ans[i-1];        b=a*a%M;b=b*alpha%M;        b=(b+a*beta)%M;        a=ans[i]=(b+gamma)%M;        if(k<=i) {            printf("%d\n",ans[k]);            return 0;        }        if(at[a]!=-1) {            T=i-at[a];  //周期            k=(k-at[a])%T;  // at[i] 是周期开始的地方，前面at[i]个数是没有在周期里的            printf("%d\n",ans[at[a]+k]);            return 0;        } else {            at[a]=i;        }    }    return 0;}