Second My Problem First HDOJ 5th Anniversary Contest 1007

Second My Problem First

Time Limit: 12000/4000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1272    Accepted Submission(s): 212

Problem Description

Give you three integers n, A and B.
Then we define S _i = A ⁱ 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 <= 10 ⁷),A and B(1 <= A, B <= 2 ³¹-1).
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
   
   
   
   

 

 

Author

WhereIsHeroFrom@HDU

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单调队列处理查询是递增的区间最小值

#include<iostream>

using namespace std;

#define N 10000000

int n;

struct QU

{

    int pos;

    __int64 value;

}Q[N];

 

int main()

{

    __int64 a,b,t,i,ans,minn,ss;

    while(scanf("%I64d%I64d%I64d",&n,&a,&b)!=EOF)

    {

        int head=1;

        int tail=0;

        t=a;

        ans=1;

        minn=a;

        for(i=1;i<=n&&a!=1;i++)

        {

            ss=a%b;

            while(head<=tail&&ss<Q[tail].value)

                tail--;

            while(head<=tail&&Q[head].pos<i-t)

                head++;

            Q[++tail].value=ss;

            Q[tail].pos=i;

            ans=(ans*Q[head].value)%b;

 

            a=(a*t)%b;

        }

        printf("%I64d/n",ans);

    }

}