hdu3706 Second My Problem First

Problem Description

Give you three integers n, A and B. 
Then we define S_i = Aⁱ mod B and T_i = Min{ S_k | i-A <= k <= i, k >= 1}
Your task is to calculate the product of T_i (1 <= i <= n) mod B.

 

Input

Each line will contain three integers n(1 <= n <= 10⁷),A and B(1 <= A, B <= 2³¹-1). 
Process to end of file.

 

Output

For each case, output the answer in a single line.

 

Sample Input

1 2 3 2 3 4 3 4 5 4 5 6 5 6 7

 

Sample Output

2 3 4 5 6

这题可以用单调队列做，存储每次的时间和大小。

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
#include<map>
#include<string>
using namespace std;
#define maxn 10000005
#define ll __int64
ll q[maxn][2];
int main()
{
    ll n,m,A,B,ans,front,rear,time,sum,timenow;
    int i,j;
    while(scanf("%I64d%I64d%I64d",&n,&A,&B)!=EOF)
    {
        front=1;rear=0;ans=1;sum=1;
        for(i=1;i<=n;i++){
            ans=(ans*A)%B;
            if(i-A<1)time=1;
            else time=i-A;
            while(front<=rear && ans<=q[rear][0]){
                rear--;
            }
            rear++;
            q[rear][0]=ans;q[rear][1]=i;
            while(front<=rear && q[front][1]<time)front++;
            sum=(sum*q[front][0])%B;
        }
        printf("%I64d
",sum%B);
    }
    return 0;
}